Download W. M. White Geochemistry Chapter 6: Aquatic Chemistry Chapter 6

W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Chapter 6: Aquatic Chemistry
6.1 Introduction
W
ater continually transforms the surface of the Earth, through interaction with the solid surface and transport of dissolved and suspended matter. Beyond that, water is essential to
life and central to human activity. Thus as a society, we are naturally very concerned with
water quality, which in essence means water chemistry. Aquatic chemistry is therefore the principal concern of many geochemists.
In this chapter, we learn how the tools of thermodynamics and kinetics are applied to water and
its dissolved constituents. We develop methods, based on the fundamental thermodynamic tools already introduced, for predicting the species present in water at equilibrium. We then examine the
interaction of solutions with solids through precipitation, dissolution, and adsorption.
Most reactions in aqueous solutions can be placed in one of the following categories:
• Acid-base, e.g., dissociation of carbonic acid:
H CO ® H+ + HCO 2
3
3
• Complexation, e.g., hydrolysis of mercury:
Hg2+ + H2 O ® Hg(OH)+ + H+
• Dissolution/Precipitation, e.g., dissolution of orthoclase:
KAlSi3 O8 + H+ + 7H2 O ® Al(OH)3 + K+ + 3H4 SiO4
• Adsorption/Desorption, e.g., adsorption of Mn on a clay surface:
∫S + Mn2+ ® ∫S–Mn
(where we are using ∫S to indicate the surface of the clay).
In this chapter, we will consider these in detail. We return to the topic of aquatic chemistry in
Chapter 13, we examine the weathering process, that is reaction of water and rock and development of soils, and the chemistry of streams and lakes.
6.2 Acid-Base Reactions
The hydrogen and hydroxide ions are often participants in all the foregoing reactions. As a result, many of these reactions are pH dependent. In order to characterize the state of an aqueous solution, e.g., to determine how much CaCO3 a solution will dissolve, the complexation state of metal
ions, or the redox state of Mn, the first step is usually to determine pH. On a larger scale, weathering of rock and precipitation of sediments depend critically on pH. Thus pH is sometimes called the
master variable in aquatic systems. We note in passing that while pH represents the hydrogen ion,
or proton concentration, the hydroxide ion concentration is easily calculated from pH since the
proton and hydroxide concentrations are related by the dissociation constant for water, i.e., by:
KW = aH + aOH -
6.1
The value of KW, like all equilibrium constants, depends on temperature, but is 10–14 at 25°C.
Arrhenius defined an acid is a substance that upon solution in water releases free protons. He defined a base is a substance that releases hydroxide ions in solution. These are useful definitions in
most cases. However, chemists generally prefer the definition of Brønstead, who defined acid and
base as proton donors and proton acceptors respectively. The strength of an acid or base is measured by its tendency to donate or accept protons. The dissociation constant for an acid or base is the
quantitative measure of this tendency and thus is a good indication of its strength. For example,
dissociation of HCl:
HCl ® H+ + Cl–
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has a dissociation constant:
K HCl =
aH + aCl = 10 3
aHCl
HCl is a strong acid because only about 3% of the HCl molecules added will remain undissociated.
The equilibrium constant for dissociation of hydrogen sulfide:
H2S ® H+ + HS–
is:
KH 2S
[H + ][HS- ]
=
= 10 -7.1
[H 2 S]
Thus H2S is a weak acid because very few of the H 2S molecules actually dissociate except at high
pH.
Metal hydroxides can either donate or accept protons, depending on pH. For example, we can
represent this in the case of aluminum as:
Al(OH) 2+ + H+ ® Al(OH)2+ + H2 O
Al(OH) 2+ + H2 O ® Al(OH) 30 + H2 O
Compounds that can either accept or donate protons are said to be amphoteric.
Metals dissolved in water are always surrounded by solvation shells. The positive charges of the
hydrogens in the surrounding water molecules are to some extent repelled by the positive charge of
the metal ion. For this reason, water molecules in the solvation shell are more likely to dissociate
and give up a proton more readily than other water molecules. Thus the concentration of such
species will affect pH.
Most protons released by an acid will complex with water molecules to form hydronium ions,
H 3O + or even H 5O +2 . However, in almost all cases we need not concern ourselves with this and
can treat the H+ ion as if it were a free species. Thus we will use [H+ ] to indicate the concentration
of H+ + H3O + + H 5O +2 +...
6.2.1 Proton Accounting, Charge Balance, and Conservation Equations
6.2.1.1 Proton Accounting
Knowing the pH of an aqueous system is the key to understanding it and predicting its behavior.
This requires a system of accounting for the H + and OH – in the system. There are several approaches to doing this. One such approach is the Proton Balance Equation (e.g., Pankow, 1991). In
this system, both H + and OH + are considered components of the system, and the proton balance
equation is written such that the concentration of all species whose genesis through reaction with water
caused the production of OH– are written on one side, and the concentration of all species whose genesis s
through reaction with water caused the production of H+ are written on the other side. Because water dissociates to form H+ and OH–, [H+] always appears on the left side (because one OH – will have been
produced for every H + produced by dissociation of water) and OH – always appears on the right
side of the proton balance equation (because one H + will have been produced for every OH – produced). The proton balance equation for pure water is thus:
[H+] = [OH– ]
6.2
Thus in pure water the concentrations of H+ and OH– are equal.
Now, consider the example of a nitric acid solution. H+ will be generated both by dissociation of
water and dissociation of nitric acid:
HNO 3 ® H + + NO 3-
Since one NO 3- ion is generated for every H+, the proton balance equation becomes:
[H +] = [OH –] + [NO 3–]
6.3
Next consider a solution of sodium chloride. Hydrogen ions may be generated by hydrolysis of
sodium:
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Na+ + H2 O ® H+ + Na(OH)
and hydroxide ions may be generated by:
Cl– + H2 O ® HCl + OH–
The proton balance equation for this reaction is:
[H+] + [HCl] = [OH– ] + [Na(OH)]
6.4
Now consider a solution of a diprotonic acid such as H 2S. H 2S can undergo 2 dissociation reactions:
H2 S ® H+ + HS–
HS– ® H+ + S2–
6.5
6.6
For every HS– ion produced by dissociation of H 2S, one H + ion would have been produced. For
every S2– ion, however, 2 H + would have been produced, one from the first dissociation and one
from the second.
The proton balance equation is thus:
[H+] = [OH– ] + [HS– ] + 2[S2–]
6.7
An alternative approach to the proton balance equation is the TOTH proton mole balance equation
used by Morel and Hering (1993). In this system, H + and H2O are always chosen as components of the
system but OH– is not. The species OH– is the algebraic sum of H2O less H+:
[OH– ] = [H2 O] – [H+]
6.8
Because aquatic chemistry almost always deals with dilute solutions, the concentration of H 2O may
be considered fixed at a mole fraction of 1, or 55.4 M. Thus in the Morel and Hering system, H2O is
made an implicit component, i.e., its presence is assumed by not written, so that equation 6.8 becomes:
[OH– ] = –[H+]
6.9
+
The variable TOTH is the total amount of component H , rather than the total of species H + . Because we create the species OH – by subtracting component H + from component H 2O, the total of
component H+ for pure water will be:
TOTH = [H+] – [OH– ]
Thus TOTH in this case is the difference between the concentrations of H+ and OH–. Now let’s consider our example of the NaCl solution. In addition to H+ and H2O, we chose Na + and Cl– as components. The species HCl is then produced by combining components H + and Cl–. The species
NaOH, however, is:
NaOH = Na+ + H2 O – H+
Thus in the proton mole balance equation, HCl counts positively and NaOH negatively:
TOTH = [H+] + [HCl] – [OH– ] – [NaOH]
6.10
Comparing equations 6.10 and 6.4, we see that the TOTH is equal to the difference between the left
and right hand sides of the proton balance equation, and that in this case TOTH = 0.
Now consider the dissolution of CO2 in water to form carbonic acid:
CO2 + H2 O ® H2 CO3
6.11
Under the right conditions of pH, this carbonic acid will dissociate to form bicarbonate ion:
H2 CO3 ® H+ + HCO 3-
6.12
In our system of accounting, we would choose CO2 as a component. The carbonic acid species
would then be made from:
H2 CO3 = H2 O + CO2
The bicarbonate ion would be made from components CO2, H2O, and H+:
HCO 3- = CO2 + H2 O - H+
Thus in the proton mole balance equation, bicarbonate ion would count negatively, so TOTH is:
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TOTH = [H+] – [OH– ] – [HCO 3- ]
6.13
We could have combined reactions 6.11 and 6.12 to write:
CO2 + H2 O ® H+ + HCO 36.2.1.3 Conservation Equations
A further constraint on the composition of a system is provided by mass balance. Acid-base reactions will not affect the total concentration of a substance. Thus regardless of reactions 6.5 and 6.6,
and any other complexation reactions, such as
Pb2+ + S2– ® PbS
the total concentration of sulfide remains constant. Thus we may write:
SS = [H2 S] + [HS– ] + [S2–] + [PbS]+ ...
We can write one mass balance, or conservation, equation for each component in solution. Of
course for components, such as Na, that form only one species, Na + in this case, the mass balance
equation is trivial. Mass balance equations are useful for those components forming more than one
species.
While the charge balance constraint is an absolute one and always holds, mass balance equations
can be trickier because other processes, such a redox, precipitation, and adsorption, can affect the
concentration of a species. We sometimes get around this problem by writing the mass balance
equation for an element, since an elemental concentration is not changed by redox processes. We
might also define our system such that it is closed to get around the other problems. Despite these
restrictions, mass balance often provides a useful additional constraint on a system.
6.2.1.2 Charge Balance
As we saw in Chapter 3, solutions are electrically neutral; that is, the number of positive and
negative charges must balance:
Si m i z i = 0
6.14
where m is the number of moles of ionic species i and z is the charge of species i. Equation 6.14 in
known as the charge balance equation and is identical to equation 3.99. This equation provides an
important constraint on the composition of a system. Notice that in some cases, the proton balance
and charge balance equations are identical (e.g., equations 6.2 and 6.7).
For each acid-base reaction an equilibrium constant expression may be written. By manipulating
these equilibrium constant expressions as well proton balance, charge balance, and mass balance
equations, it is possible to predict the pH of any solution. In natural systems where there are many
species present, however, solving these equations can be a complex task indeed. An important step
in their solution is to decide which reactions have an insignificant effect on pH and neglect them.
6.2.3 The Carbonate System
We now turn our attention to carbonate. Water at the surface of the Earth inevitably contains dissolved CO2, either as a result of equilibration with the atmosphere or because of respiration by organisms. CO 2 reacts with water to form carbonic acid:
CO2 + H2 O ® H2 CO3
6.15
Some of the carbonic acid dissociates to form bicarbonate and hydrogen ions:
H2 CO3 ® H+ + HCO 3-
6.16
Some of the bicarbonate will dissociate to an additional hydrogen ion and a carbonate ion:
HCO 3- ® H+ + CO 32 -
6.17
We can write three equilibrium constant expressions for these reactions:
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a H 2CO3
ƒ CO2
a H+ a HCO3K1 =
a H 2CO3
a H+ a CO32K2 =
a HCO3–
6.18
K sp =
6.19
6.20
The equilibrium constants for these reactions are given in Table 6.1 as a function of temperature.
In the above series of reactions, we have simplified things somewhat and have assumed that dissolved CO2 reacts completely with water to form H2CO 3. This is actually not the case, and much of
the dissolved CO2 will actually be present as distinct molecular species, CO2(aq ). Thus reaction 6.18
actually consists of the two reactions:
CO2(g) + H2 O ® CO2(aq )
6.15a
Example 6.1. Proton, Mass, and Charge Balance Equations for Na2CO3 Solution
Write the proton, proton mass balance, and charge balance, and carbonate conservation
equations for a solution prepared by dissolving Na2CO3 in water. Assume that NaCO3
dissociates completely and that the system is closed.
Answer: We begin with the proton balance equation. From the dissociation of water we
have:
[H+] = [OH– ]
2–
In addition to this, hydroxide ions will also be generated by reaction between CO 3 and water:
CO 23 – + H2 O ® HCO 3– + OH–
HCO 3– + H2 O ® H2 CO3 + OH–
and
Since for each HCO 3– formed, one OH– must have formed and for each H 2CO3 present two OH –
must have formed, the proton balance equation is:
[H+] + [HCO 3– ] + 2[H2 CO3 ] = [OH– ]
6.21
+
Choosing the carbonate and sodium ions as components (in addition to H and H 2O), the
bicarbonate and carbonic acid species are combinations of carbonate and one and two,
respectively, protons, so that the proton mole balance equation is:
TOTH = [H+] + [HCO 3– ] + 2[H2 CO3 ] – [OH– ]
+
6.22
+
Alternatively, we could chose CO2 (in addition to H , H 2O and Na ) as our forth component,
rather than carbonate ion. In this case, the three carbonate species are made from components
as follows:
H2 CO3 = H2 O + CO2
HCO 3– = H2 O + CO2 – H+
CO 23 – = H2 O + CO2 – 2H+
In this case, the proton mole balance equation is:
TOTH = [H+] - [HCO 3– ] - 2[CO 23 – ] – [OH– ]
6.22a
Here we see that TOTH depends on how we define our components.
The charge balance equation is:
[H+] + [Na+] = [OH– ] + [HCO 3– ] + 2[CO 23 – ]
6.23
The conservation equation for carbonate species is:
SCO3 = [CO 23 – ] + [HCO 3– ] + [H2 CO3 ]
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Table 6.1. Equilibrium Constants for the Carbonate System
T (°C)
0
5
10
15
20
25
30
45
60
80
90
pKCO2
pK1
pK2
pK cal
pKarag
K CaHCO 3+ *
1.11
1.19
1.27
1.34
1.41
1.47
1.52
1.67
1.78
1.90
1.94
6.58
6.52
6.46
6.42
6.38
6.35
6.33
6.29
6.29
6.34
6.38
10.63
10.55
10.49
10.43
10.38
10.33
10.29
10.20
10.14
10.13
10.14
8.38
8.39
8.41
8.43
8.45
8.48
8.51
8.62
8.76
8.99
9.12
8.22
8.24
8.26
8.28
8.31
8.34
8.37
8.49
8.64
8.88
9.02
–0.82
–0.90
–0.97
–1.02
–1.07
–1.11
-1.14
–1.19
–1.23
–1.28
–1.31
pK CaCO 30 †
–3.13
–3.13
–3.13
–3.15
–3.18
–3.22
-3.27
–3.45
–3.65
–3.92
–4.05
*KCaHCO 3+ = aCaHCO 3+ /(aCa 2+aHCO 3- )
† K
0
0
CaCO 3 = a CaCO 3 /(aCa 2+a CO
23
)
CO2(aq ) + H2 O ® H2 CO3
0
6.15b
Common pH
range in nature
log a
10.33
2–
The equilibrium for the second reaction
–
H2CO3
CO
HCO
3
3
favors CO2(aq ). However, it is analyti-2
cally difficult to distinguish between the
species CO2(aq ) and H 2CO 3. For this
+ EP
H
reason, CO2(aq ) is often combined with
OH–
-4
H 2CO3 when representing the aqueous
6.4
species. The combined total concentration of CO2(aq ) + H 2CO3 is sometimes
-6
written as H2CO *3 . We will write it
1
3
5
7
9
11
13
simply as H2CO3.
pH
The importance of the carbonate system is that by dissociating and provid- Figure 6.1. Activities of different species in the carbonate
-2
ing hydrogen ions to solution, or asso- system as a function of pH assuming SCO2 = 10 . After
ciating and taking up free hydrogen Drever (1988).
ions, it controls the pH of many natural waters. Example 6.2 shows that pure water in equilibrium
with atmospheric CO2 will be slightly acidic. The production of free H+ ions as a result of the solution of CO2 and dissociation of carbonic acid plays an extremely important role is weathering.
Natural waters with a pH equal to or greater than 7 must contain cations other than H+ .
Ground waters may not be in equilibrium with the atmosphere, but will nonetheless contain
some dissolved CO 2. Because of respiration of organisms in soil (mainly plant roots & bacteria)
through which they pass before penetrating deeper, ground waters often contain more CO2 than
water in equilibrium with the atmosphere. In addition, calcite and other carbonates are extremely
common minerals in soils and in sedimentary, metamorphic, and altered igneous rocks. Ground
waters will tend to approach equilibrium with calcite by either dissolving it or precipitating it:
CaCO3 ® Ca2+ + CO 32 -
6.25
Carbonate ions produced in this way will associate with hydrogen ions to form bicarbonate as in
reaction 6.17 above, increasing the pH of the solution. Water containing high concentrations of
calcium (and magnesium) carbonate is referred to as ‘hard water’; such waters are generally somewhat alkaline.
Now suppose we have a known activity of all carbonate species in solution, say for example 10-2:
aH 2CO 3 + aHCO - + aCO 2- = SCO2 = 10 -2
3
3
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Example 6.2. pH of Water in Equilibrium with the Atmosphere
What is the pH of water in equilibrium with the atmospheric CO2 at 25°C, assuming ideal behavior and no other dissolved solids or gases present? The partial pressure of CO2 in the atmosphere is
3.5 ¥ 10-4.
Answer: In this case, the proton balance and charge balance equations are identical:
[H + ] = [OH- ] + [HCO-3 ] + 2[CO 23- ]
6.26
We might guess that the pH of this solution will be less than 7 (i.e., [H+ ] ≥ 10-7). Under those circumstances, the concentrations of the hydroxyl and carbonate ions will be much lower than that of
the hydrogen and bicarbonate ions. Assuming we can neglect them, our equation then becomes
simply:
[H + ] @ [HCO-3 ]
6.26a
We can combine equations 6.18 and 6.19 to obtain an expression for bicarbonate ion in terms of the
partial pressure of CO2:
(
)
[HCO -3 ] = K1KCO 2 PCO 2 / [H + ]
Substituting this into 6.26a and rearranging, we have:
[H + ]2 @ K1K 2 KCO 2 PCO 2
6.27
Taking the negative log of this expression and again rearranging, we obtain:
pH @
- log K1 - log KCO 2 - log PCO 2
2
Substituting values from Table 6.1, we calculate pH = 5.64. Looking at Figure 6.1, we can be assured that our assumption that carbonate and hydroxyl ion abundances are valid. Indeed, an exact
solution using the Solver in Excel™ differs from the approximate one by less than 0.0001 pH units.
From this, and the dissociation constants, we can calculate the amount of each species present as a
function of pH and temperature. For example, we can use the equilibrium constant expressions to
obtain substitutions for the carbonic acid and carbonate ion activities in equation 6.28 that are functions of bicarbonate ion activity and pH. We then solve equation 6.28 to obtain an expression for
the activity of the bicarbonate ion as a function of total CO2 and hydrogen ion activity:
aHCO - =
3
SCO2
aH +
K
+1+ 2
K1
aH +
6.29
Similar equations may be found for carbonic acid and carbonate ion. Carrying out these calculations at various pH, we can construct the graph shown in Figure 6.1. In this figure, we see that carbonic acid is the dominant species at low pH, bicarbonate at intermediate pH, and carbonate at
high pH.
Particularly simple relationships occur when the activities of two species are equal. For example,
choosing pH to be 6.35 (at 25°C) we can rearrange equation 6.19 and substitute to obtain:
aH + aH 2CO 3 10 -6.35
=
=
=1
aHCO - 10 -6.35
K1
3
6.30
The activity of the carbonate ion will be very low at this pH. If we ignore it, equ. 6.28 becomes:
aH 2CO 3 + aHCO - @ 10 -2
3
According to equ. 6.30, the activities of both species are equal, so each must be 5 ¥ 10-3.
Now consider the point where the hydrogen ion concentration equals the bicarbonate ion concentration. At this point, the concentration of the carbonate ion is extremely low, and there is exactly enough H+ to convert all HCO 3- to H2CO 3. From the perspective of the proton balance then,
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Example 6.3. pH of a Solution with Fixed Total Carbonate Concentration
A groundwater moving through soil into a deep aquifer acquires a total dissolved CO2 concentration of 10–2 M. Assuming the water does not exchange with surrounding rock, ideal behavior and
no other dissolved solids or gases, what is the pH of the water?
Answer: In this case, our charge and proton balance equations are the same as in Example 6.2,
i.e., equation 6.26. Since the solution does not exchange with surrounding rock, it can be considered a closed system and we can write the following mass balance equation:
SCO2 = [ H2 CO3 ] + [ HCO3– ] + [CO32 - ] = 10 -2
6.31
Simultaneously solving the change balance and mass balance equations, and using equilibrium
constant expressions to eliminate carbonate and OH species, we obtain:
[H + ]4 + K1[H + ]3 + {K 2 K1 - K W - K1SCO 2}[H 2 + ] - {K w + 2 K 2 SCO 2}K1[H + ] - K 2 K1K W = 0
We might again guess that the concentration of the carbonate ion will be very low, and that we can
therefore neglect all terms in which K 2 occurs. We might also guess that pH will be acidic so that
[H+] ›› [OH–], and therefore that we can neglect terms containing KW. Our equation becomes:
K1-1[H + ]2 + [H + ] = SCO2
Solving this quadratic, we find that pH = 4.18.
the HCO 3- concentration is equivalent to the same concentration of H 2CO 3. This point, labeled EP
on Figure 6.1, is called the CO2 equivalence point. In a similar way, the point where the carbonic
acid and carbonate ion concentrations are equal is called the bicarbonate equivalence point, and that
where bicarbonate and hydroxyl concentrations are equal is called the carbonate equivalence point.
The pH of these equivalence points depends, among other things, on the S CO 2 concentration.
The exact concentrations of carbonate species depends on total carbonate concentration as well as
the concentration of other ions in solution. Thus the distribution shown in Figure 6.1 is unique to
the conditions specified (CT = 10-2, no other ions present). Nevertheless, the distribution will be
qualitatively similar for other conditions.
6.2.4 Conservative and Non-Conservative Ions
We can divide dissolved ions into conservative and non-conservative ones. The conservative ions
are those whose concentrations are not affected by changes in pH, temperature, and pressure, assuming no precipitation or dissolution. In natural waters, the principal conservative ions are Na + , K + ,
Ca 2+, Mg2+, Cl–, SO 24 and NO 3- . These ions are conservative because they are fully dissociated from
their conjugate acids and bases over the normal range of pH of natural waters. Non-conservative
ions are those that will undergo association and dissociation reactions in this pH range. These in–
clude the proton, hydroxide ion, and carbonate species as well as B(OH) 4 , H3SiO4 , HS –, NH 4OH,
phosphate species, and many organic anions. Virtually all the non-conservative species are anions,
the two principle exceptions being H + and NH 4OH (which dissociates to form N H +4 at high pH).
Variations in the concentrations of non-conservative ions result from reactions between them, and
these reactions can occur in the absence of precipitation or dissolution. For example, reaction of
the carbonate and hydrogen ion to form bicarbonate will affect the concentrations of all three ions.
Of course, if the system is at equilibrium, this reaction will not occur in the absence of an external
disturbance, such as a change in temperature.
6.2.5 Total Alkalinity and Carbonate Alkalinity
Alkalinity is a measure of acid-neutralizing capacity of a solution and is defined as the sum of the
concentration (in equivalents) of bases that are titratable with strong acid. Mathematically, we define
alkalinity as the negative of TOTH when the components are the principal components of the solution at the
CO 2 equivalence point. The acidity can be defined as the negative of alkalinity, and hence equal to
TOTH.
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Chapter 6: Aquatic Chemistry
As a first example, let’s consider a solution containing a fixed total dissolved concentration of
CaCO3. At the CO2 equivalence point, H 2CO3 (or CO2(aq )) is the principal carbonate species, so we
chose our components as H+, H2O, CO2, and Ca2+. Species HCO 3- and CO 32 - are made by combining these components as follows:
HCO 3- = H2 O + CO2 – H+
CO 32 - = H2 O + CO2 – 2H+
The proton mole balance equation is then:
TOTH = [H+] – [HCO 3– ] – 2[CO 23 – ] – [OH– ]
6.32
The alkalinity is then:
Alk = –TOTH = –[H+] + [HCO 3– ] + 2[CO 23 – ] + [OH– ]
+
3]
2[CO 32 - ]
6.33
–
This sum, –[H ] + [HCO
+
– [OH ], is called the carbonate alkalinity. In this example,
carbonate alkalinity and alkalinity are equal since there are no other ions in solution. To avoid
confusion with carbonate alkalinity, alkalinity is sometimes called total alkalinity.
Now let’s consider a somewhat more complex example, a solution that contains H3SiO -4 ,
B(OH) -4 , and HS–, and phosphoric acid (H3PO 4 and dissociated species). This solution is similar to
seawater. The corresponding components, the species actually present at the CO2 equivalence
point, are H4SiO4, B(OH)3, H2S, and H2PO 4– . Our expression for alkalinity is then (assuming negligible concentrations of S2_ and PO 3–
4 ):
Alk = –[H+] + [HCO 3– ] + 2[CO 23 – ] + [OH– ]
+ [B(OH) -4 ] + [H3 SiO -4 ] + [HS- ] + [H2 PO 4– ] + 2[HPO 24 – ]
6.34
An analytical definition of alkalinity is that it is the quantity of acid that must be added to the solution to bring the pH to the CO2 equivalence point.
We can also express alkalinity in terms of conservative and non-conservative ions. The charge
balance equation, equation 6.14, could be written as:
Scations (in equivalents) – S anions (in equivalents) = 0
6.35
This can the be expanded to:
Sconserv. cations - Sconserv. anions + Snon-conserv. cations - Snon-conserv. anions = 0
(all in units of equivalents)* . Rearranging, we have:
Sconserv. cations - Sconserv. anions = -Snon-conserv. cations + Snon-conserv. anions
6.36
The right hand side of equation 6.32 is equal to the alkalinity. Hence we may write:
Alk=Sconserv. cations - Sconserv. anions = -Snon-conserv. cations + Snon-conserv. anions 6.37
This equation emphasizes an important point. The difference of the sum of conservative anions and
cations is clearly a conservative property, i.e., they cannot be changed except by the addition or removal of components. Since alkalinity is equal to this difference, alkalinity is also a conservative
quantity (i.e., independent of pH, pressure and temperature). Thus total alkalinity is conservative,
though concentrations of individual species are not.
* One equivalent of a species is defined as the number of moles multiplied by the charge of the species. Thus one
equivalent of CO 32 - is equal to 0.5 moles of CO 32 - , but one equivalent of Cl– is equal to 1 mole of Cl– . For an acid
or base, an equivalent is the number moles of the substance divided by the number of hydrogen or hydroxide
ions that can be potentially produced by dissociation of the substance. Thus there are 2 equivalents per mo le of
H2 CO3 , but 1 equivalent per mole of Na(OH).
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Example 6.4. Calculating Alkalinity of Spring Water
Calculate the alkalinity of spring water from Thonon,France, Anions
mM
Cations mM
whose analysis is given at right (this is the same analysis as
HCO 3
5.436
Ca 2+
2.475
in Problem 3.8).
2SO 4
0.146
Mg2+
0.663
Answer: We can use equation 6.34 to calculate alkalinity.
NO 30.226
K+
0.036
All the ions listed here are conservative with the exception
Cl –
0.231
Na +
0.223
of HCO 3 . To calculate alkalinity, we first need to convert
the molar concentrationsto equivalents; we do so by multiplying the concentrationof each species by
its charge. We find the sumof conservative anion concentrationsto be 0.749 meq (milliequivalents),
and that of the conservative cation concentrationsto be 6.535 meq. The alkalinity is the difference,
5.786 meq.
6.2.5.1 Alkalinity Determination and Titration Curves
If the concentrations of all major conservative ions in a solution are known, the alkalinity can be
simply calculated from equation 6.35. It is often useful, however, to determine this independently.
This is done, as the definition of alkalinity suggests, through titration. Titration is the process of
progressively adding a strong acid or base to a solution until a specified pH, known as an endpoint, is reached. In the case of the determination of alkalinity, this end point is the CO2 equivalence point.
Consider a solution containing a certain concentration of sodium bicarbonate (Na2CO3). Because
the carbonate ion can act as a proton acceptor, NaCO3 is a base. We can determine both the alkalinity and the total carbonate concentration of this solution by titrating with a strong acid, such as
HCl. Let’s examine the chemistry behind this procedure.
For clarity, we make several simplifying assumptions. First, we assume ideal behavior. Second,
we assume the system is closed, so that all components are conserved, except for [H+] and [Cl-],
which we progressively add. Third, we assume that the volume of our Na2CO3 solution is sufficiently large and our HCl sufficiently concentrated that there is no significant dilution of the original solution. Finally, we assume both Na2CO 3 and HCl dissociate completely.
The charge balance equation during the titration is:
[Na+] + [H+] = [Cl– ] + [HCO 3- ] + 2[CO 23 – ] + [OH– ]
6.38
–
Since the Cl concentration is conservative, it will be equal to the total amount of HCl added. Into
equation 6.38, we can substitute the following:
[HCO13- ] =
K1[H 2 CO 3 ]
K K [H CO ]
K
6.39a [CO 23 - ] = 1 2 + 22 3 6.39b and [OH- ] = W+ 6.39c
+
[H ]
[H ]
[H ]
Doing so and rearranging yields:
[Cl - ] = [Na + ] + [H + ] -
K1[H 2 CO 3 ] K1K 2 [H 2 CO 3 ] KW
- +
[H + ]
[H + ]2
[H ]
6.40
We may also write a conservation equation for carbonate species, which is the same as equation
6.24 in Example 6.1. Substituting equations 6.37a and 6.37b into 6.24 and rearranging, we have:
[H 2 CO 3 ] =
SCO2
K
KK
1 + +1 + 1 + 22
[H ] [H ]
6.41
Substituting this expression into 6.40, we obtain:
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Chapter 6: Aquatic Chemistry
Ï
K1K 2 ¸ KW
6.42
ýÌK1 K1K 2 Ó
[H + ] þ [H + ]
+
[H ] + K1 +
[H + ]
From stiochiometry, we also know that S CO 2 = [Na+]. From this equation we can construct a plot
SCO2
[Cl - ] = [Na + ] + [H + ] -
showing how many moles of HCl we must add to achieve a certain pH. We can also use equation
6.39 and similar ones expressing the bicarbonate and carbonate ions as functions of pH to plot the
change in the carbonate speciation during the titration. Figure 6.2 shows such a plot for a 0.005 M
Na2CO3 solution. There are two regions where pH changes rapidly with small additions of HCl.
These are the two end-points of the titration. Comparing the titration curve with the speciation
curves, we see that the two end-points correspond to the CO2 and bicarbonate equivalence points.
An analytical definition of alkalinity is its acid neutralizing capacity when the end-point of the titration
is the CO2 equivalence point (Morel and Hering, 1993). We had previously defined alkalinity as the
negative of TOTH when the principal components are those at the CO2 equivalence point. Let’s
now show that these definitions are equivalent.
Our TOTH expression, written in terms of components at the CO2 equivalence point, is identical
to 6.22a:
TOTH = [H+] – [HCO 3- ] – 2[CO 23 – ] – [OH– ]
(6.32)
and the charge balance equation (before any HCl is added) is:
[Na+] + [H+] = [HCO 3- ] + 2[CO 23 – ] + [OH– ]
(6.14)
Combining the two we have:
TOTH = –[Na+]
Since the alkalinity is the negative of TOTH, it follows that (before the addition of HCl):
Alk = [Na+]
6.43
We obtain exactly the same result from equation 6.37. It is easy to show that after titrating to the
CO 2 equivalence point, the alkalinity is 0. The change in alkalinity is thus equal to the number of
equivalents, or moles, of H+ we have added to the solution. Since at the end point, [H+] = [HCO 3- ]
and the concentrations of CO 32 - and OH– are negligible, our charge balance equation, 6.38, reduces
to:
16
14
12
10
8
6
4
2
0
2
E.P. 2
0
E.P. 1
H2CO3
-1
HCO3-
-2
-3
-
H+
4
OH
2-
CO 3
6
pH
-4
Log a
Comparing this with 6.41, we see
that the alkalinity is equal to the
amount of HCl added. In the example in Figure 6.2, the equivalence point occurs after the addition of 10 ml of 1M HCl, or a total
of 0.01 moles of Cl–. (Notice that
since at the end points, small
additions of acid result in large
changes in pH, we do not have to
determine pH particularly accurately during the titration for an
accurate determination of alkalinity.)
So the alkalinity is 0.01
equivalents. This is exactly the
answer we obtain from 6.41 for 1
liter of 0.005 M Na2CO3 since
there are 2 moles of Na + for each
mole of Na2CO 3.
ml 1M HCl
[Na+] = [Cl– ]
-5
-6
8
-7
10
Figure 6.2. Titration curve (solid red line) for a one liter 0.005 M
Na2CO3 solution titrated with 1M HCl. Left axis shows the number of ml of HCl to be added to obtain a given pH. Also shown
are the concentrations of carbonate species, H+, and OH(dashed black lines, right axis gives scale).
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6.44
Thus total carbonate is obtained by titrating to the
bicarbonate equivalence point (knowing the pH of
the end-point allows us to determine the SCO2
exactly; however neglecting the [OH–] term in 6.42
results in only a 1% error in the example shown). In
Figure 6.2, this occurs after the addition of 5 ml 1M
HCl.
6.2.6 Buffer Intensity
The carbonate system is a good example of a pH
buffer. We define the buffer intensity of a solution as
the inverse of change in pH per amount of strong
base (or acid) added:
b∫
dC B
dC A
=dpH
dpH
100
Buffer Intensity, meq/l-pH
+
3.5
l
Kao .
SCO2 = [Cl ] + [OH ]
–
b An-
By assuming that the concentration of H +
contributes negligibly to charge balance, it is also
easily shown (Problem 6.2) that at the bicarbonate
equivalence point:
=
10
–
10
bC
O2
b PC
aCO
3 (s)
1
b
CT =
0.1
7
10 –3
8
pH
9
6.45
Figure 6.3. Buffer intensity as a function
of pH for several ideal natural systems:
where C B and C A are the concentrations, in equiva- bC T fixed total dissolved CO2, bPCO water in
lents, of strong base or acid respectively.
The equilibrium with atmospheric CO2, bCaCO3(s)
greater the buffer capacity of a solution, the less water in equilibrium with calcite, and bAnchange in its pH as an acid or base is added. The Kaol. water in equilibrium with anorthite
buffer capacity of a solution can be calculated by dif- and kaolinite. After Stumm and Morgan
ferentiation of the equation relating base (or acid) (1996).
concentration to pH, as is illustrated in Example 6.5.
In nature, a pH buffer acts to control pH within a narrow range as H+ ions are added or removed
from solution by other reactions. To understand how this works, imagine a solution containing
2carbonic acid, CO3 , HCO 3 , and H + ions in equilibrium concentrations. Now imagine that additional H + ions are added (for example, by addition of rain water containing HNO 3). In order for
the right hand side of equation 6.19 to remain equal to K 1 despite an increase in the activity of H +
(which it must at constant temperature and pressure), the bicarbonate activity must decrease and the
carbonic acid activity increase. It is apparent then that reaction 6.16 must be driven to the left,
taking free hydrogen ions from solution, hence driving the pH back toward its original value.
Similarly, reaction 6.20, the dissociation of bicarbonate, will also be driven to the left, increasing the
bicarbonate concentration and decreasing the hydrogen and carbonate ion concentrations.
The buffer capacity of the carbonate system depends strongly on pH and also on the concentration of the carbonate species and the concentration of other ions in solution. In pure water containing no other ions and only carbonate in amounts in equilibrium with the atmosphere, the buffering
capacity is negligible near neutral pH, as is shown in Figure 6.4. Natural solutions, however, can
have substantial buffering capacity. Figure 6.3 illustrates three other examples of natural pH buffers. "Hard water" is an example of a water with a substantial buffering capacity due to the presence
of dissolved carbonates. As we shall see, how adversely lakes and streams are impacted by “acid
rain” depends on their buffering intensity.
2
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Chapter 6: Aquatic Chemistry
Example 6.5. Calculating Buffer Intensity
How will pH change for given addition of a strong base such as NaOH for a solution of pure
water in equilibrium with atmospheric CO2? Calculate the buffer intensity of this solution as a
function of pH. Assume that NaOH completely dissociates and ideal behavior.
Answer: We want to begin with the charge balance equation in this case because it relates the
two quantities of interest in this case, [Na+] and [H+]. The charge balance equation is the same as
in Example 6.1:
[Na + ] + [H+ ] = [OH - ] + [HCO 3- ] + 2[CO 23 - ]
(6.23)
+
Since Na is a conservative ion, its concentration will depend only on the amount of NaOH
added, so that CB = [Na+]. Subsituting this into equation 6.14 and rearranging, we have:
C B = [OH–] + [HCO3–] + 2[CO32–] – [H+] C B == [OH - ] + [HCO 3- ] + 2[CO 23 - ] - [H+ ]
6.46
We can now use the equilibrium constant relations to substitute for the first three terms of the right
hand side of 6.46 and obtain:
KW + K1Ksp PCO 2
+
[H ]
+2
K 2 K1Ksp PCO 2
+ 2
[H ]
Using the relation pH = - log [H+ ] to replace [H+ ] in
this equation with pH, we have:
CB =
KW + K1Ksp PCO 2
+2
- pH
10
K 2 K1Ksp PCO 2
10
-2 pH
4
- [H+ ]
- 10 - pH
2
log b, eq l-1 pH-1
CB =
0
-2
Now differentiating with respect to pH, we obtain:
-4
dC B
= b = ln 10{(Kw + K1K sp PCO 2 )10 pH
dpH
6.47
-6
+4K 2 K1K sp PCO 2 10 2 pH + 10 - pH }
2
4
6
pH
8
10
12
Figure 6.4 shows a plot of this equation using the Figure 6.4.
Buffer capacity of a
values in Table 6.1. Buffer intensity is negligible in carbonate solution in equilibrium with
neutral to slightly acidic conditions, but increases atmospheric CO2.
rapidly with pH.
6.3 Complexation
Ions in solution often associate with other ions, forming new species called complexes. Complex
formation is important because it affects the solubility and reactivity of ions, as we will see in the
following section. In some cases, complex formation is an intermediate step in the precipitation
process. In other cases, ions form stable, soluble complexes that greatly enhance the solubility of
the one or both of the ions.
Complexation is usually described in terms of a central ion, generally a metal, and ions or molecules that surround, or coordinate, it, referred to as ligands. Perhaps the simplest and most common
complexes are those formed between metals and water or its dissociation products. We learned in
Section 3.7 that ions in aqueous solutions are surrounded by a solvation shell. The solvation shell
(Figure 3.10) consists of water molecules, typically 6, though fewer in some cases, loosely bound to
the ion through electrostatic forces. This solvation shell is referred to as an aquo complex. Water
molecules are the ligands in aquo-complexes. Aquo-complexes are ubiquitous: all charged species
will have a solvation shell. Truly “free ions” do not exist: ions not otherwise complexed (“free
ions”) are in reality associated with surrounding water molecules and hence actually aquo-
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complexes. However, the existence of this type
of complex is implicitly accounted for through
the activity coefficient and not usually explicitly
considered. Nevertheless, it is important to bear
in mind that since all ions are complexed in some
–
+
+
–
way to begin with, every complexation reaction
in aqueous solution is essentially a ligand
exchange reaction.
Beyond aquo complexes, we can distinguish
two types of complexes:
a. Solvation Shell Contact
b. Shared Solvation Shell
• Ion pairs, where ions of opposite charge associate with one and other through electrostatic
attraction, yet each ion retains part or all of it’s
solvation sphere. Figure 6.5 illustrates two possibilities: one where the two solvation spheres are
merely in contact, the other where the water
+ –
molecules are shared between the two solvation
spheres. Ion pairs are also called outer sphere
complexes.
• Complexes (senso stricto), where the two ions
c. Ion Contact
are in contact and a bond forms between them
that is at least partly covalent in nature (Figure Figure 6.5. Illustration of ion pair and complex
6.5c). These are often called inner sphere com- formation. Two types of ion pairs can be enviplexes.
sioned: (a)solvation shell contact and (b) solvation
shell sharing. Ion pairs are sometimes referred to
6.3.1 Stability Constants
as outer sphere complexes. In formation of true
In its simplest form, the reaction for the for- complexes, ions are in contact (c) and there is
mation of an ion pair or complex between a some degree of covalent bonding between them.
metal cation M and an anion or ligand L may be These are some times referred to as inner sphere
written as:
complexes.
mM+ + lL– ® MmLl
Na+
As with any other reaction, we may define an
I
NaOH
ns 2+
equilibrium constant as:
+
al Io
Ion Pairs
Complex (Senso Stricto)
For example, the
reaction:
6.48
aMm aLl
equilibrium
constant
for
the
Zn(H2 O) 62 + + OH– ® Zn(H2 O)5 OH+ + H2 O
aZn ( H O ) ( OH ) +
2
5
is:
K1 =
aZn ( H O ) 2+ aOH 2
6
(we omit the activity of water because we assume, as
usual, that it is 1). As we noted, however, the aquocomplex is generally not explicitly expressed, so this
same reaction would more often be written as:
Zn2+ + OH– ® ZnOH+
Valance State
K=
Fe
FeOH
Met
o
u
s
Aq Al 3+ Al(OH)
l Ion
n
a
3+
t
e
Fe
III
M
Fe(OH)n
4+ Pu(OH)
roxo 2–
d
Pu
n
y
IV Zr 4+ Zn(OH)n CO2(OH)– H CO3
II
aM mL l
V
s
CrO3 (OH)
l Ion
a
2–
t
e
–
CrO 4
M
VI SO3(OH) SO42–
Oxo
MnO (OH)MnO–4
VII ClO43(OH)
VIII OsO4
1
3 5 7
9 11
pH
13
Figure 6.6. Predominant aquo, hydroxo, and
oxo complexes as a function of pH and valance
state. After Stumm and Morgan (1996).
and the equilibrium constant as:
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Chapter 6: Aquatic Chemistry
K1 =
aZn ( OH ) +
aZn 2+ aOH -
Equilibrium constants for complex formation reactions are often referred to as stability constants,
since their value is an indication of the stability of the complex, and often denoted by b. Thus for
the reaction above, b1 and K1 are synonymous. By convention, stability constants are written so as the
complex appears in the numerator (i.e., as a product of the reaction).
The zinc ion might associate with a second hydroxyl:
The equilibrium constant is:
ZnOH– + OH– ® Zn(OH)2
aZn ( OH ) 2
K2 =
aZn ( OH ) + aOH -
Here, however, the notation for the stability constant and the equilibrium constant differs. Whereas
K 2 refers to the reaction above, b2 refers to the reaction:
Zn2+ + 2OH– ® Zn(OH)2
aZn ( OH ) 2
b2 =
= K1K 2
2
aZn 2+ aOH
-
Hence:
Finally, the notation *K and *b are sometimes used for reactions in which the complexation reactions are written so as the hydrogen ion occurs as a product, for example:
Zn2+ +H2 O ® ZnOH+ + H+
aZn ( OH ) + aH +
*
K1 =
aZn 2+
and:
*K1 is then related to K1 as:
*K1 = K1 KW = *b1
where KW is the water dissociation constant (10-14).
We can define apparent equilibrium and stability constants, where the molar concentrations are
used in place of activity. Indeed, as in other aspects of geochemistry, apparent equilibrium
constants are more commonly encountered than true ones.
The equilibrium constant may in turn be related to the Gibbs Free Energy of the reaction, as in
equation 3.86. Interestingly, the free energy changes involved in complexation reactions result
largely from entropy changes. Indeed, the enthalpy changes of many complexation reactions are
unfavorable, and the reaction proceeds only because of large positive entropy changes. These entropy changes result from the displacement of water molecules from the solvation shell.
The link between the equilibrium constant and the free energy change is particularly important
and useful in complexation reactions because it is in most instances difficult to determine the concentrations of individual complexes analytically. Thus our knowledge of chemical speciation in
natural waters derived largely from predictions based on equilibrium thermodynamics.
6.3.2 Water-Related Complexes
Let’s further consider the types of complexes typically found in aqueous solution. Ferric iron, for
example, can form a Fe(H2 O) 63+ complex. The positive charge of the central ion tends to repel hydrogens in the water molecules, so that water molecules in these aquo-complexes are more readily
hydrolyzed than otherwise. Thus these aquo-complexes can act as weak acids. For example:
Fe(H O) 3+ ® Fe(H O) (OH)2+ + H+ ® Fe(H O) (OH) + + 2H+ ®
2
6
2
5
2
4
2
Fe(H2 O)3 (OH) 30 + 3H+ ® Fe(H2 O)2 (OH) 4– + 4H+
6.49
As reaction 6.49 suggests, equilibrium between these hydroxo -complexes depends strongly on pH.
The repulsion between the central metal ion and protons in water molecules of the solvation shell
will increase with decreasing diameter of the central ion (decreasing distance between the protons
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Chapter 6: Aquatic Chemistry
and the metal) and with increasing charge of the central ion. For highly charged species, the repulsion of the central ion sufficiently strong that all hydrogens are repelled and it is surrounded only
2by oxygens. Such complexes, for example, MnO 4– and CrO 4 , are known as oxo-complexes.
Intermediate types in which the central ion is surrounded, or coordinated, by both oxygens and hydroxyls are also possible, for example MnO3(OH) and CrO3(OH)–, and are known as hydroxo-oxo
complexes. Figure 6.6 summarizes the predominance of aquo, hydroxo, hydro-oxo, and oxo complexes as a function of pH and valance state. For most natural waters, metals in valance states I and
II will be present as “free ions”, i.e., aquo complexes, valance III metals will be present as aquo
and hydroxo complexes, those with higher charge will present as oxo-complexes.
Polynuclear hydroxo- and oxo-complexes, containing two or more metal ions, are also possible,
for example:
OH–
Pb2+
–1
OH
Pb2+ OH–
3+
Fe
Fe3+
2+
2+
Mn — OH — Mn
Pb2+
OH–
OH–1
OH–
Pb2+
As one might expect, the extent to which such polymeric species form depends on the concentration of the metal ion concentration: they become increasingly common as concentration increases.
Most highly charged (3+ and higher oxidation states) metal ions are highly insoluble in aqueous
solution. This is due in part to the readiness with which they form hydroxo-complexes, which can
in turn be related to the dissociation of surrounding water molecules as a result of their high
charge. When such ions are present at high concentration, formation of polymeric species such as
those above quickly follows formation of the hydroxo complex. At sufficient concentration,
formation of these polymeric species leads to the formation of colloids and ultimately to
precipitation. In this sense, these polymeric species can be viewed as intermediate products of
precipitation reactions.
Interestingly enough, however, the tendency of metal ions to hydrolyze decreases with
concentration. The reason for this is the effect of the dissociation reaction on pH. For example,
increasing the concentration of dissolved copper decreases the pH, which in turn tends to drive the
Example 6.6. Complexation of Pb
Assuming a Pb concentration of 10–8 M and an equilibrium constant for the reaction:
Pb2 + + H2 O ® PbOH+ + H+
of 10-7. 7, calculate the fraction of Pb that will be present as PbOH+ from pH 6 to 9.
Answer: In addition to the above reaction, we also need the conservation equation for Pb:
SPb = Pb2 + + PbOH+
1
The equilibrium constant expression is:
[ PbOH + ][H + ]
K=
Pb 2 +
Solving the conservation equation for Pb 2+ and
substituting into the equilibrium constant
expression, weo obtain:
(SPb – [PbOH+])K = [PbOH+][H]
With some rearranging, we eventually obtain the
following expression:
[PbOH+ ] SPb =
K
[H ] + K
fraction PbOH+
0.8
0.6
0.4
0.2
0
+
6
7
pH
8
9
The result is illustrated in Figure 6.7. Below pH 6, Figure 6.7. Fraction of Pb complexed as
virtually all Pb is present as Pb2 + , above pH 9, PbOH + as a function of pH.
virtually all Pb is present as PbOH+.
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hydrolysis reaction to the left. To understand this, consider the following reaction:
for which the apparent equilibrium constant is K app = 10–8. We can express the
fraction of copper present as CuOH+, aCuOH +
as:
a=
[CuOH+ ]
K
= +
Cu T
[H ] + K
6.50
pH, –log (a)
Cu2+ + H2 O ® CuOH+ + H+
7
6
5
4
3
2
1
pH
–log (a)
where CuT is the total dissolved copper. At
constant pH, the fraction of Cu complexed
0
0.5
1.0
1.5
2.0
CuT 10-4 M
is constant. However, for a solution with a
fixed amount of Cu ion dissolved, we can Figure 6.8. pH and –log a, as a function of total copalso write a proton balance equation:
per concentration in aqueous solution. a is the frac[H +] = [CuOH +] + [OH –]
tion of copper present as the hydroxo-complex.
and a mass balance equation. Combining
these with the equilibrium constant expression, we can calculate both a and pH as a function of CuT
(Problem 6.9). When we do this, we see that as CuT increases, both pH and a decrease, as is
demonstrated in Figure 6.8.
6.3.3 Other Complexes
When non-metals are present in solution, as they would inevitably be in natural waters, then
other complexes are possible. In this respect, we can divide the elements into four classes (Table
6.2, Figure 6.9). The first is the non-metals, which form anions or anion groups. The second group
is the “A-type” or “hard” metals. These metals, listed in Table 6.2, have spherically symmetric,
inert-gas type outer electron configurations. Their electron shells are not readily deformed by
electric fields and can be viewed as “hard spheres”. Metals in this group preferentially forms com2
3
2
plexes with fluorine and ligands having oxygen as the donor atoms (e.g., OH –, CO 3 , PO 4 , SO 4 ).
Stability of the complexes formed by these metals increases with charge to radius ratio. Thus the alkalis form only weak, unstable complexes, while elements such as Zr 4+ form very strong, stable
complexes (e.g., with fluorine). In complexes formed by A-type metals, anions and cations are
bound primarily by electrostatic forces, i.e., ionic-type bonds. The A-type elements correspond
approximately to the lithophile elements of Goldschmidt’s classification presented in Chapter 7.
The third group is the B-type, or “soft”, metal ions. Their electron sheaths are not spherically
symmetric and are readily deformed by the electrical fields of other ions (hence the term soft). They
preferentially form complexes with bases having S, I, Br, Cl, or N (such as ammonia; not nitrate) as
the donor atom. Bonding between the metal
and ligand(s) is primarily covalent and is Table 6.2. Classification of Metal Ions
comparatively strong. Thus Pb form strong
complexes with Cl– and S2–. Many of the A-Type Metals
+
+
+
+
+
2+
2+
2+
2+
complexes formed by these elements are quite Li , Na , K , Rb , Cs , Be , Mg , Ca , Sr ,
2+
3+
3+
3+
4+
4+
4+
insoluble. The B-type elements consist primarily Ba , Al , Sc , Y , REE, Ti , Si , Zr ,
4+
4+
5+
5+
6+
of the “chalcophile elements”, a term we will Hf , Th , Nb , Ta , U
define in the next chapter.
B-Type Metals
The first series transition metals form the Cu 2+, Ag+, Au+, Tl+, Ga+, Zn2+, Cd2+,
fourth group, and correspond largely to the Hg 2+, Pb2+, Sn2+, Tl3+, Au3+, In3+, Bi3+
siderophile elements (see Chapter 7).
Their Transition Metal Ions
electron sheaths are not spherically symmetric, 2+
, Cr2+, Mn2+, Fe2+, Co2+, Ni2+,
but they are not so readily polarizable as the B- V 2+
3+
3+
+
3+
3+
3+
type metals. On the whole, however, their Cu , Ti , V , Cr , Mn , Fe , Co
From Stumm and Morgan (1996).
232
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
He
H
Li Be
B
C
N O
F
Ne
Na Mg
Al
Si
P
S
Cl
Ar
Zn Ga Ge
As Se
Br
Kr
Cd
In Sn
Sb
Te
I
Xe
Au Hg
Tl Pb
Bi Po
At
Rd
K Ca
Sc
Ti
Rb Sr
Y
Zr Nb Mo Tc Ru Rh Pd Ag
Cs
La
Hf
Ba
V Cr
Ta W
Mn Fe
Re Os
Co
Ni Cu
Ir
Pt
Fr Ra Ac
Pr
Nd Pm Sm
Th Pa
U Np Pu
La Ce
Ac
A-Metals
Eu Gd
Transition
Metals
Tb Dy Ho Er Tm Yb
B Metals
Lu
Ligand
Formers
Figure 6.9. Classification of the elements with respect to complex formation in aqueous solution.
Log Kspsulfide
Log K1
complex-forming behavior is similar to that of the B-type metals.
Among the transition metals, the sequence of complex stability is Mn 2+<Fe2+<Co2+<Ni2+
<Cu2+>Zn2+, a sequence known as the Irving-Williams Series. This is illustrated in Figure 6.10. In
that figure, all the sulfate complexes have approximately the same stability, a reflection of the
predominately electrostatic bonding between sulfate and metal. Pronounced differences are
observed for organic ligands. The figure demonstrates an interesting feature of organic ligands: although the absolute value of stability
E
complexes varies from ligand to ligand,
12
the relative affinity of ligands having the
F
same donor atom for these metals is
2S
30
10
always
similar.
Ethylandiamine
Organic molecules can often have more
H
E
than one functional group and hence can
8
F
coordinate a metal at several positions, a
E
E
process called chelation. Such ligands
G
H
6
F
F
are called multi-dentate and organic comG
H
H
20
Glycinate G
pounds having these properties are reG
H
4 H E
ferred to as chelators or chelating agents.
Oxalate
G
J
We will explore this topic in greater deJ
J
J
J
J
F
SO42tail in the Chapter 14.
2 E
The kinetics of complex formation is
quite fast in most cases, so that equilib0
10
rium can be assumed. There are excepMn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+
tions, however. As we noted earlier in
Figure 6.10. Stability constants for transition metal
this section, all complexation reactions
sulfate and organic complexes and their sulfide
are ligand exchange reactions: water
solubility constants, illustrating the Irving-Williams
playing the role of ligand in “free ions”.
series. From Strumm and Morgan (1996).
233
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
The rate at which complexes form is thus governed to a fair degree by the rate at which water
molecules are replaced in the hydration sphere.
6.3.4 Complexation in Fresh Waters
Where only one metal is involved, the complexation calculations are straightforward, as exemplified in Example 6.6. Natural waters, however, contain many ions. The most abundant of these are
Na + , K + , Mg 2+, Ca 2+, Cl–, SO 24 - , HCO 3– , CO 32 - , and there are many possible complexes between
them as well as with H + and OH –. To calculate the speciation state of such solutions an iterative
approach is required.
The calculation would be done as follows.
First, we need the
concentrations, activity coefficients, and stability constants (or apparent stability constants) for all
species. Then we assume all ions are present as free ions and calculate the concentrations of the
various possible complexes on this basis. In this pass, we need only consider the major ions (we
can easily understand why with an example: formation of PbCl+ when the concentration of Pb is 108
or less and the abundance of Cl– is 10-4 or more will have an insignificant affect on the free ion Cl
concentration). We then iterate the calculation, starting with the free ion concentrations corrected
for abundances of complexes we calculated in the previous iteration. This process is repeated until
two successive iterations produce the same result. Although it sounds difficult, such calculations
typically converge within 2 to 4 iterations. Example 6.7 shows how this is done. Once free ion
concentrations of the major ligands are known, the speciation of trace metals may be calculated.
As Example 6.7 demonstrates, the major metals in fresh waters are present mainly as free ions
(aquo complexes), as are the three most common anions, chloride, sulfate, and bicarbonate. The alkali and alkaline earth trace elements are also largely uncomplexed. Co2+, Ni2+, Zn2+, Ag+, and Cd2+
are roughly 50% complexed. The remaining metals are present as primarily as complexes. B, V ,
Cr, Mo, and Re, as well as As and Se are present as anionic oxo-complexes. Other metals are
usually present as hydroxide, carbonate, chloride, or organic complexes.
Under reducing
conditions, HS– and S2- complexes are important. In organic-rich waters such as swamps, organic
complexes can be predominant. We will discuss organic complexes in more detail in Chapter 14,
and speciation in seawater in Chapter 15.
6.4 Dissolution and Precipitation Reactions
6.4.1 Calcium Carbonate in Ground and Surface Waters
Calcium carbonate is an extremely common component of sedimentary rocks and is found
present in weathered igneous and metamorphic rocks. It is also a common constituent of many
soils. Water passing through such soils and rocks will precipitate or dissolve calcite until
equilibrium is achieved. This process has a strong influence on carbonate concentrations, hardness,
and pH as well as dissolved calcium ion concentrations. Let’s examine calcite solubility in more
detail.
The solubility product of calcite is:
Ksp -cal = aCa 2+ aCO 2-
6.52
3
This can be combined with equations 6.18–20 to obtain the calcium concentration water in
equilibrium with calcite as a function of PCO 2:
[Ca 2 + ] = PCO 2
K1K sp -cal K sp -CO 2
2
- 2
K 2g Ca 2+ g HCO
- [HCO 3 ]
6.53
3
In a solution in equilibrium with calcite and a CO2 gas phase and containing no other dissolved
species, it is easy to modify equation 6.51 so that the calcium ion concentration is a function of PCO 2
only. A glance at Figure 6.1 shows that we can neglect OH– , H+ , and CO 32 - if the final pH is less
than about 9. The charge balance equation in this case reduces to:
234
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Example 6.7. Speciation in Fresh Water
of Stream Water (mM)
Using the water and stability constants given in the Analysis
+
0.32 Cl–
0. 22
adjacent tables, calculate the activities of the major Na
+
2–
K
0.
06
SO
0. 12
species in this water.
4
2+
Mg
0.
18
SCO
1.0
Answer: The first two problems we need to address
2
2+
are the nature of the carbonate species and activity Ca
0. 36 pH
8.0
coefficients. At this pH, we can see from Log Stability Constants
Figure 6.1 that bicarbonate will be the
OH –
HCO 3– CO 23 – SO 2–
Cl–
4
dominant carbonate species. Making the initial
+
14
6.35 10.33
1.99
–
assumption that all carbonate is bicarbonate, H
+
–
-0.25
1.27
1.06
–
we can calculate ionic strength and activity Na
–
–
–
0.96
–
coefficients using the Debye-Hückel Law K +
(equation 3.74). These are shown in the table Mg 2+
2.56
1.16
3.4
2.36
–
below. Having the activity coefficients, we can Ca2+
1.15
1.26
3.2
2.31
–
calculate the approximate abundance of the
carbonate ion by assuming all carbonate is bicarbonate:
[CO 23 - ] =
g HCO 3- SCO 2
baH + g CO 32-
where b is the stability constant for the complexation reaction:
CO 23 - + H+ ® HCO -3
The “corrected” bicarbonate ion
is then calculated as:
[HCO -3 ] = SCO2 - [CO 23 - ]
The result confirms our initial first
order assumption that all
carbonate
is
present
as
bicarbonate.
Using the concentrations and
Ion Activities: Iteration 1
free ion
free ion
H+
Na +
K+
Mg 2+
Ca2+
OH –
HCO 3–
1¥10-06 9.17¥10-04
—
2.12¥10-05
—
1.62¥10-07
—
—
-08
2.0310-06
5.09¥10
1¥10-08
3.03¥10-04
5.69¥10-05
1.39¥10-04
2.77¥10-04 4.29¥10-09
5.08¥10-06
CO 23 –
SO 2–
4
6.79¥10-07 1.62¥10-04
9.46¥10-04 1.75¥10-10
2.51¥10-08 6.27¥10-07
—
9.33¥10-08
1.65¥10-06 6.10¥10-06
2.07¥10-06 1.08¥10-05
stability constants given as well as
Ion Activities: Iteration 2
the activity coefficients we calcufree ion OH –
HCO 3–
CO 23 –
SO 2–
4
lated, we can them make a first
-06
-04
-06
free
ion
pass at calculating the concentra1¥10
9.12¥10
1.12¥10 1.65¥10-04
+
-08
-05
tions of the complexes.
For H
—
1¥10
2.06¥10
9.12¥10-04 1.58¥10-10
example, MgCO3 is calculated as
Na +
—
3.03¥10-04
1.57¥10-07 2.35¥10-08 5.64¥10-07
aMgCO 3 = b MgCO 3 aMg 2+ aCO 2K+
—
—
—
5.69¥10-05
8.40¥10-08
3
2+
The results of this first iteration Mg
1.40¥10-04 5.03¥10-08 1.84¥10-06 1.45¥10-06 5.14¥10-06
2+
are shown in a matrix. Chlorine Ca
2.80¥10-04 3.92¥10-09 4.64¥10-06 1.83¥10-06 9.16¥10-06
does not complex with any of the
major ions to any significant degree, so we can neglect it in our calculations.
We then correct the free ion activities by subtracting the
% Free Ion
activities of the complexes they form. Thus for example, the Na +
99.76% Cl–
100%
2+
corrected free ion activity of Mg is calculated as:
+
2–
K
99.85% SO 4
91.7%
corr
ini
2+
–
aMg
2+ = a
2+ - a
- - a
- - a MgCO - a MgSO
Mg
94.29%
99.3%
HCO 3
Mg
MgOH
MgHCO 3
3
4
2+
2
–
94.71% CO 3
25.3%
We then repeat the calculation of the activities of the Ca
complexes using these corrected free ion activities. A second matrix shows the results of this
second iteration. A third table shows the percent of each ion present as a free ion (aquo complex).
In fresh waters such as this one, most of the metals are present as free ions, the alkaline earths being
5% complexed by sulfate and carbonate.
235
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Ca2+ (mmol/l)
4
4
3
3
2
2
1
1
0
-4
-3
-2
Log PCO2
Superstaturated
A
C
00
-1
Understaturated
B
0.02
0.04
0.06
0.08 0.10
PCO2
Figure 6.11. Concentration of calcium ion in equilibrium with calcite at 25°C and
1 atm as a function of PCO 2. From Drever (1988).
2[Ca 2+ ] = [HCO -3 ]
4
6.54
Substituting this into 6.53, we obtain:
[Ca ] = PCO 2
K1K sp -cal K sp -CO 2
2
2+ 2
]
4K 2g Ca 2+ g HCO
- [Ca
6.55
3
1/ 3
or
ÏÔ
K1K sp -cal K sp -CO 2 ¸Ô
[Ca ] = ÌPCO 2
ý
2
4K 2g Ca 2+ g HCO
- Ô
ÔÓ
3 þ
2+
6.56
3
Ca2+ (mmool/l)
2+
2
+
Na
=0
-3 m
10
=
+
Na
3m
10¥
=5
+
Na
There are two interesting aspects to this equation.
2
First, the calcium ion concentration, and therefore cal10- m
=
+
1
cite solubility, increases with increasing P CO2. This
Na
might seem counter-intuitive at first, as one might think
that that increasing PCO2 should produce an increase the
carbonate ion concentration and therefore drive the reaction toward precipitation. However, increasing PCO2
0
0 0.02 0.04 0.06 0.08 0.10
decreases pH, which decreases CO 2-3 concentration,
and therefore drives the reaction towards dissolution.
PCO2
Second, calcium ion concentration varies with the onethird power of PCO2 (Figure 6.11). Because of this non- Figure 6.12. Concentration of calcium ion
linearity, mixing of two solutions, both of which are in a solution containing Na2CO3 in
saturated in Ca2+ with respect to calcite, can result in equilibrium with calcite at 25°C and 1 atm
the mixture being undersaturated with respect to Ca2+. as a function of PCO 2 and sodium conFor example, consider the mixing of stream and ground centration. From Drever (1988).
water.
Stream water is in equilibrium with the
atmosphere for which PCO2 is 10--3.5 . On the other hand, P CO2 in soils is often as high as 10-2. So mixing between calcite-saturated groundwater and calcite-saturated surface water would produce a solution that is undersaturated with calcite.
Calcite solubility is also affected by the presence of other ions in solution. Figure 6.12 illustrates
the effect of sodium ion concentration on calcite solubility.
Equation 6.56 describes calcite solubility for a system open to exchange with gaseous CO 2. For a
PCO2 of 10-3.5 (i.e., the atmosphere), this equation yields a calcium concentration of 1.39 mM. Water
in pores and fractures in rocks does not exchange with a gas phase. Example 6.8 shows that under
those circumstances, less calcite will dissolve; in the case of PCO2 initial = 10-2, calcite saturation is
236
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Example 6.8. Calcite Solubility in a Closed System
Suppose ground water initially equilibrates with a PCO2 of 10-2 and thereafter is closed to gas
exchange, so that there is a fixed SCO2 ini tial. The water then equilibrates with calcite until
saturation is reached. What will be the final concentration of calcium in the water.
Assume ideal behavior and an initial calcium concentration of 0.
Answer: Since the system is closed, a conservation equation is a good place to start. We
can write the following conservation equation for total carbonate:
SCO 2 = SCO 2initial + SCO 2 from calcite
Since dissolution of one mole of calcite adds one mole of
equation may be rewritten as:
SCO2 for each mole of Ca2+, this
SCO2 = SCO2 initial + [Ca2+ ]
Neglecting the contribution of the carbonate ion to total carbonate, this equation becomes:
[H2 CO3 ] + [HCO 3- ] = ([H2 CO3 ]initial ) + [Ca2+ ]
6.57
where ([H2CO3]initial ) denotes that amount of H 2CO3 calculated from equation 6.21 for
equilibrium with CO2 gas; in this case a partial pressure of 10-2. This can be rearranged to
obtain:
([H2 CO3 ]initial ) = [H2 CO3 ] + [HCO 3- ] – [Ca2+ ]
6.58
Further constraints are provided by the three carbonate equilibrium product expressions
(6.21–6.23) as well as the solubility product for calcite (6.50), and the charge balance equation.
We assume a final pH less than 9 and no other ions present, so the charge balance equation
reduces to equation 6.54. From equation 6.18 and the value of K CO 2 in Table 6.1, [H2CO3]initial
= 10-2 ¥ 10--1.47 M. Dividing equation 6.19 by 6.20 yields:
[HCO -3 ]2
K1
=
K 2 [H 2 CO 3 ][CO 32 - ]
Then substituting equations 6.52, 6.54, and 6.58 gives:
K1
4[Ca 2 + ]3
=
K 2 K sp -cal [H 2 CO 3 ]initial - [Ca 2 + ]
{
}
Into this equation we substitute PCO K CO = [ H 2CO 3 ] and rearrange to obtain:
2
[Ca 2+ ]3 +
2
K1Kcal KCO 2
K1Kcal
[Ca 2+ ] ( PCO 2initial ) = 0
4K 2
4K 2
6.59
This is a cubic that is readily solved for [Ca2+]. For an initial PCO 2 of 10-2, we calculate a
calcium concentration of 0.334 mM
reached at only 0.33 mM, or about a fourth as much. The difference is illustrated in Figure 6.13,
which is a plot of [HCO 3– ] vs. pH. Systems in equilibrium with constant PCO2 (open systems) evolve
along straight lines on this plot and ultimately reach calcite saturation at higher pH and lower
[HCO 3– ] (and [Ca2+]) than closed systems that initially equilibrate with the same PCO2.
6.4..2 Solubility of Mg
There are a number of compounds that can precipitate from Mg-bearing aqueous solutions, including brucite (Mg((OH)2), magnesite (MgCO3), dolomite (CaMg(CO3)2), as well as hydrated carbonates such as hydromagnesite (MgCO3(OH)2. 3H 2O). The stability of these compounds may be
described by the following reactions:
Mg(OH)2 ® Mg2+ + 2OH–
Kbru = 10–11.6
237
6.60
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
MgCO 3 ® Mg 2+ + CO 23 CaMg(CO 3 ) 2 ® Mg
2+
+ Ca
2+
Kmag = 10–7.5
+ 2 CO
23
6.61
–17
Kdol = 10
6.62
(The solubility of dolomite is poorly known; values for this equilibrium constant vary between
10-16.5 and 10–20.)
We can use these reactions and their equilibrium constants, together with the reactions for the carbonate system (equ. 6.15-6.17) to construct predominance diagrams for Mg-bearing solutions in
equilibrium with these phases. For example, for reaction 6.60, we may derive the following relationship (assuming ideal solution):
-2
= 5.0
al
Assuming the solid is a pure phase, the
term on the left of each equation is then just
-log[Mg2+]. We can use these equations to
construct stability, or predominance diagrams
in a manner similar to that used to construct
pe-pH predominance diagrams (Chapter 4).
For example, on a plot of log[Mg2+] vs. pH,
the predominance boundary between
2+
Mg aq
and brucite plots as a line with a
slope of –2 and an intercept of +16.4. Figure 6.14 shows two predominance diagrams derived from these equations. Brucite is the stable phase at high pH and low
PCO2.
-2
Mg2+
aq
-6
6
8
10
pH
12
atm
ite E
quili
b
r ium
2
10 –
atm
= 5.0
10 –
3
=
2
2
2
O
6
=
al
ni ti
2
3 tm
0– a
1
pH
, pH intial = 5.3
8
7
Figure 6.13. Comparison of the evolution of systems
with constant PCO2 (open systems) with those closed to
gas exchange. After Stumm and Morgan (1996) and
Deines et al. (1974).
0
b
-2
Magnesite
-4
-4
=
=
PC
5
a
Magnesite
P CO
-5
Brucite
log [Mg2+]
0
ial
int
, pH intial
m
t
a
2
PC
+ 2
-4
0–
1
=
O
6.64
1
PC
1
[H ] K1K 2 + [H+ ] K 2 + 1
O
2
nt
2i
O
C
P
Log PCO2
+
log (HCO–3)
[MgCO 3 ]
log
= pK mag + log SCO2
[Mg 2 + ]
=
ia l
atm
O
-3
1
0–
Calc
inti
, pH
PC
where we use the notation pK = –log (K).
For reaction 6.61, we can derive the following:
atm
6.63
1
[Mg(OH) 2 ]
= pK bru - 2 pK W
[Mg 2 + ]
+2 pH = -16.4 + 2 pH
10 –
log
Mg2+
aq
-6
14
-8
Brucite
6
8
10
pH
12
14
Figure 6.14. Predominance diagrams for Mg-bearing phases in equilibrium with
aqueous solution. Total CO2 is fixed at 10-2.5 M in 6.14a. The concentration of Mg 2+
is fixed at 10-4 M in 6.14b. After Stumm and Morgan (1996).
238
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
0
-4
CaMg(CO3 )2 + Ca+ ® Mg2+ + 2CaCO3
Dolomite
Magnesite
-2
Log PCO2
Virtually all natural solutions will contain dissolve calcium as well as magnesium. This being the case, we must
also consider the stability of dolomite. We can construct
similar predominance diagrams for these systems but we
must add an additional variable, namely the Ca2+ concentration. To describe the relative stability of dolomite and
calcite, it is more convenient to express the solubility of
dolomite as:
-6
Calcite
Brucite
because the reaction contains calcite as well as dolomite.
-8
Since this reaction can be constructed by subtracting 2
-4
-2
0
2
4
times the calcite dissolution (equ. 6.50) from the dolomite
2+
2+
dissolution (6.60), the equilibrium constant for this reaction
can be calculated from:
Figure 6.15. Stability of magnesite,
dolomite, calcite, and brucite in equiK
K = dol
librium with a Mg- and Ca-bearing
2
Kcal
aqueous solution.
Figure 6.15 illustrates the stability of magne0
site, dolomite, brucite and calcite as a func2+
2+
tion of PCO2 and the Ca /Mg concentration
ratio. Whether any of these phases are stable
-1
relative to a Mg 2+-bearing solution depends
on the Mg 2+ concentration, which is not
specified in the graph.
-2
6.4.3 Solubility of SiO2
Silicon is the most common elements on the
Earth's surface after oxygen. Its concentration
in solution plays an important role in determining how weathering will proceed.
The dissolution of silica may be represented
by the reaction:
SiO2(qtz ) + 2H2 O ® H4 SiO4(( aq )
Log activity SSiO2
log[Ca ]/[Mg ]
Amorphous silica
-3
-4
2
6.65
H2SiO42-
Total
H4SiO4
4
6
pH
8
H3SiO4–
10
12
14
The equilibrium constant expression is simFigure 6.16. Log activity of dissolved silica in equiply:
librium with quartz and amorphous silica (dashed
-4
K qtz = aH 4 SiO 4 = 10 at 25°C
6.66
line) as a function of pH. After Drever (1988).
This is to say, water is saturated with respect
to quartz when the concentration of H2SiO4 is 10-4 moles per kilogram, or about 7.8 ppm by weight
SiO2.
However, there are some complicating factors. First, precipitation of quartz seems to be strongly
kinetically inhibited. Equilibrium is more likely to be achieved with amorphous silica, the equilibrium constant for which is 2 ¥ 10-3 (~115 ppm). Second, H4SiO4 is a weak acid and undergoes successive dissociation with increasing pH:
H 4SiO 4 ® H 3SiO -4 + H+
and
H 3SiO -4 ® H 2 SiO 24 - + H+
239
K1 =
K2 =
aH
3 SiO 4
aH
aH
4
aH +
SiO 4-
22 SiO 4
aH
aH +
3 SiO 4
= 10 -9.9
= 10 -11.7
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Example 6.8. Constructing Stability Diagrams
Using the equilibrium constant data below as well as from Table 6.1, construct a stability diagram showing the stablity of FeS, siderite (FeCO3(S )), and Fe(OH)2 as a function of total sulfide concentration and pH assuming SCO 2 = 5 ¥ 10-2 M, Fe2+ = 10-6 M, and ideal behavior. Neglect any S2–.
FeS( S) + H+ ® Fe2 + + HS–
FeCO3 (S) + H+ ® Fe2 + + HCO 3–
Fe(OH)2 (S ) + 2H+ ® Fe2 + + 2H2O
H2 S( aq ) ® H+ + HS–
KFeS = 10- 4. 2
KFeCO 3 = 10- 0. 1
K Fe( OH )2 = 101 2. 9
KS = 10- 7
6.67
6.68
6.69
6.70
Answer: Let’s first consider reaction 6.67 above. Our first step is to set up an equation that
describes the concentration of HS– as a function of pH. From the conservation of sulfur, we have:
SS = [H2 S] +[ HS– ]
From the equilibrium constant expression for the dissociation of H2S, we may substitute:
[H 2S] =
[H + ][HS– ]
KS
SS =
and obtain:
[HS– ] = SS
Solving for [HS–] we have:
[H + ][HS– ]
+ [HS– ]
KS
6.71
KS
K S + [H + ]
We substitute this into the FeS solubility product and solving for SS we have:
KFeS [H+ ](KS + [H+ ])
[Fe 2 + ]KS
log(SS) = log(KS + [H+]) –pH – log([Fe2 +]) + pKFeS – pKS
SS =
or in log form:
6.72
This plots as line ① on our SS vs. pH stability diagram (Figure 6.17). The area above the line is the
region where FeS is stable.
Next, let’s consider reaction 6.68, the solubility of siderite. We need an equation describing the
concentration of HCO 3– as a function of pH, which is equation 6.29. Substituting this into the
siderite solubility product we have:
KFeCO 3 =
or in log form:
[Fe 2+ ]SCO 2
Ï[H+ ]
K ¸
+ 1 + +2 ý[H + ]
Ì
[H ] þ
Ó K1
0 = pH + log [Fe 2+] + log SCO2 – log
K
[H+]
+ 1 + +2 – pK FeCO3
K1
[H ]
6.73
An approximate solution may be found by assuming HCO 3– = SCO 2 , which yields pH = 8.20. An
exact solution requires an indirect method. Using the Solver in Microsoft Excel™, we obtain pH =
8.21, very close to our approximate solution (Solver uses a succession of “intelligent” guesses to
find solutions to equations, such as 6.73, that have no direction solution). Thus siderite will
precipitate when the pH is greater than 8.21 and, not surprisingly, this is independent of SS. The
boundary between the Fe2+ and FeCO3 field is then a vertical line (line ②) at pH = 8.21.
Now let’s consider the solubility of ferrous iron hydroxide, The condition for precipitation of
Fe(OH)2 is described by the equation:
pH =
pKFe(OH) 2 - log[Fe 2+ ]
2
= 9.45
6.74
Thus FeOH2 will not precipitate until pH reaches 9.5, i.e., above the point where FeCO3 precipitates, so there is no boundary between a Fe(OH)2 phase and a Fe2+ -bearing solution.
Next we need equations that describe the reactions between solid phases. A reaction between
FeCO 3 and Fe(OH)2 can be obtained by subtracting reaction 6.69 from reaction 6.68. The corresponding equilibrium constant is obtained by dividing 6.69 by 6.68:
240
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
FeCO3 + 2H2 O ® (FeOH)2 + H+ + HCO 3–
From this we derive:
pH = log [HCO 3– ] + 13.02
KFeCO 3 / K Fe( OH )2 = 10- 13.0 2
6.75
–
3,
We can again substitute equation 6.29 for HCO
but it is difficult solve the resulting equation for
pH. We can proceed in one of two ways. First, we ca we can obtain an approximate solution, by
simply assuming all carbonate is bicarbonate, in which case, we obtain pH = 10.72. An alternative
is to use a tool such as the Solver in Excel™, which yields an exact solution of pH =10.5. Not
surprisingly, our approximate solution is less accurate than in the previous case because at this
high pH, the carbonate ion makes up a significant fraction of the total carbonate (Figure 6.1). The
boundary between the FeCO3 and Fe(OH)2 fields is thus a vertical line at pH = 10.5 (line ③).
The boundary between FeS and FeCO3 is found by subtracting reaction 6.68 from 6.67 and
dividing the corresponding equilibrium constants:
FeS + HCO 3– ® FeCO3 + HS–
KFeS/ KFeCO 3 = 10- 4. 1
6.76
Substituting 6.71 and 6.29 into the corresponding equilibrium constant expression, we have:
log[HS- ] - log[HCO 3- ] =
0
log SS + pKS - log(K S + [H+ ]) Ï[H + ]
K ¸
log SCO 2 + log Ì
+ 1 + +2 ý = -4.1
[H ] þ
Ó K1
log SS = - pKS - log(K S + 10 - pH ) +
Ï10 - pH
K ¸
log SCO 2 - log Ì
+ 1 + -2pH ý - 4.1
10 þ
Ó K1
From the equilibrium constant equation we
obtain an expression of SS as a function of
pH:
FeS
➄
-6
➃
2+
Fe aq
-8
This plots as line ➃ on our diagram. Finally,
the reaction between FeS and Fe(OH)2 is obtained by subtracting reaction 6.69 from reaction 6.67:
FeS + 2H2 O ® Fe(OH)2 + H+ + HS–
KFeS/ K Fe( O H) 2 = 10- 17.1
①
-4
log SS
Solving for SS and substituting 10-pH for [H+],
we have:
-2
-10
➁
4
6
8
FeCO3 ➂ Fe(OH)2
10
12
14
Figure 6.17. Stability diagram showing the stable
solid Fe-bearing phases in equilibrium with a
solution containing 10 -6 M Fe2+ and 5 ¥ 10-3 M
SCO2.
log SS = pH –pKS + log(KS + 10- pH)
This plots as line ➄ on our diagram, which is now complete.
The total dissolved silica concentration will be the sum of H 4SiO4, H3SiO -4 , and H2SiO -4 2 . Assuming activity coefficients of unity, the concentration of dissolved silica is then:
K
K K ¸Ô
ÔÏ
[SiO 2 ]T = [H 4SiO 4 ]Ì1 + 1 + 12 2 ý
aH + Ôþ
ÔÓ aH +
6.77
From equation 6.77, we would expect silica solubility to be pH dependent. This dependence is illustrated in Figure 6.16.
We could have defined the second dissociation reaction as:
H 4SiO 4 ® H 2 SiO 24 - + 2 H+
241
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
-1
Ca2+
log [Me]
-2
-3
-4
Fe3+
-5
-6
Fe2+
Al3+
2
4
Cu2+
6
Zn2+
Mg2+
Cd2+
8
pH
Ag+
10
12
Figure 6.18. Solubility of metal hydroxides as a function of pH. After Stumm and Morgan (1981).
In which case, the equilibrium constant would be:
K =
*
2
aH
22 SiO 4
aH
4
aH2 +
SiO 4-
= 10 -9.9 ¥ 10 -11.7 = 10 -21.6
and equation 6.77 would have been:
ÏÔ
K
K* ¸Ô
[SiO 2 ]T = [H 4SiO 4 ]Ì1 + 1 + 2 2 ý
ÓÔ aH + aH + þÔ
6.77a
The concentration of SiO2T we calculate in this way would, of course, be identical. The point is, reactions and their corresponding equilibrium constants can be expressed in various ways, and we
need to be alert to this.
6.4.4 Solubility of Al(OH)3 and Other Hydroxides
Al(OH)3(s) + 3H+ ® Al3+ + 3H2 O
a 3+
K gib = Al3 = 10 -8.1
6.78
aH +
However, a complication arises from hydrolyzation of the aluminum, which occurs in solutions
that are not highly acidic (hydrolyzation is typical of many of the highly charged, ≥3, metal
ions):
242
3+
-2 Al
Log activity of dissolved Al species
The hydroxide is the least soluble salt of many
metals. Therefore, it is the solubility of their hydroxides that controls the solubility of these metals in natural waters. These are shown in Figure
6.18. Since these dissolution reactions involve
OH– , they are pH dependent, and the slope of
the solubility curve depends on the valence of the
metal (e.g., -3 for Fe 3+ , -2 for Fe 2+ , -1 for Ag 3+).
Let’s consider in more detail the solubility of
gibbsite, the hydroxide of aluminum.
Dissolution of gibbsite (Al(OH)3) can be described by the reaction:
Al(OH)2+
Supersaturated
-4
-6
Al(OH)–4
Al(OH)+2
Al(OH)03
-8
-10
2
4
6
pH
8
10
Figure 6.19. Log activity of dissolved aluminum species and total Al (solid red line) in
equilibrium with gibbsite as a function of
pH
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
K1 =
Al3+ + H2 O ® Al(OH)2+ + H+
Al
3+
Al
Al
3+
3+
+ 2 H 2 O ® Al(OH) + 2 H
+
2
+ 3H 2 O ® Al(OH) + 3H
0
3
K3 =
+
+ 4H 2 O ® Al(OH) + 4H
13
K2 =
+
K4 =
+
a Al ( OH ) 2+ aH +
a Al 3+
a Al ( OH ) + aH2 +
2
a Al 3+
a Al ( OH ) 0 aH3 +
3
a Al 3+
a Al ( OH )1- aH4 +
3
a Al 3+
= 10 -5
6.79
= 10 -9.3
6.80
= 10 -16
6.81
= 10 -22
6.82
Neglecting activity coefficients, the total dissolved aluminum concentration is given by:
ÏÔ
K
K
K
K ¸Ô
[ Al 3+ ]T = [ Al 3+ ]Ì1 + 1 + 2 2 + 3 3 + 4 4 ý
ÔÓ aH + aH + aH + aH + Ôþ
6.83
Figure 6.18 shows the concentrations of the various aluminum species and total aluminum as a
function of pH. The solubility of Al is low except at low and high pH, and that as pH increases
Al3+ becomes increasingly hydrolyzed. Also note that where positively charged species dominate,
e.g., Al 3+, solubility increases with decreasing pH; where negatively charged species dominate,
solubility increases with increasing pH. Minimum solubility occurs in the range of pH of natural
waters (which is just as well because it has some toxicity).
Equation 6.72 is a general result that can be applied to metals other that undergo hydrolyzation
reactions, when the reactions are expressed in the same form as those above. A general form of this equation would thus be:
¸Ô
ÏÔ
K
K
K
[M]T = [M z ]Ì1 + 1 + 2 2 + 3 3 + Ký
Ôþ
ÔÓ aH + aH + aH +
6.84
Thus iron and other metals show a pH dependence similar to Al. For Fe, the relevant equilibrium
constants are:
Fe(OH) 2 + + H + ® Fe 3 + + H 2 O
Fe(OH) +2 + 2 H + ® Fe 3 + + 2 H 2 O
log K = 5.7
6.84b
3+
+
+
Fe(OH) 4 + 4H ® Fe + 4H 2 O
log K = 21.6
6.84c
+
3+
Fe(OH) 3( s ) + 3H ® Fe + 3H 2 O
log K = 3.2
6.84d
-2
amorph. Fe(OH)3
log [SFe]
-4
-6
Fe(OH)+2
-8
Fe3+
-10
Fe(OH)2+
Fe(OH)–4
0
2
4
6
Using these equilibrium constants, the solubility of amorphous goethite as a function of
pH is readily calculated and is shown in
Figure 6.20.
pH
8
10
12
14
Figure 6.20. Solubility of goethite as a function of
pH. Solubility of individual Fe-hydroxide species
shown as dashed lines.
243
6.4.5 Dissolution of Silicates and
Related Minerals
The concentrations of Al and Si will usually not be controlled by equilibrium with
quartz and gibbsite, rather by equilibrium
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
e
llit
y
h
e
rop olinit ite
y
P Ka bbs
Gi
-8
-10
2
6
pH
log aAl(total)
-4
-6
-8
-10 a
-3
H4SiO4 = 10
2
4
6
pH
-8
-10
aH4SiO4 = 10-5
4
-6
log aAl(total)
-6
lite
l
y
h
rop bsite
y
P ib ite
G olin
Ka
-4
8
10
4
6
pH
8
10
-4
site ite
b
b
Gi phyll
ro te
Py aolini
K
8
aH4SiO4 = 10-4
2
log aAl(total)
log aAl(total)
-4
-6
-8
-10
10
aH4SiO4 = 10-2
2
4
6
pH
e
bsit
b
i
G ite
olin te
Ka phylli
ro
Py
8
10
Figure 6.21. Total dissolved Al activity in equilibrium with gibbsite, pyrophyllite, and kaolinite as
a function of pH at different dissolved silica activities. After Drever (1988).
with other silicates. An example of this is shown in Figure 6.21, which shows the concentration of
dissolved Al in equilibrium with gibbsite, kaolinite and pyrophyllite at four different activities of
dissolved silica. Only at the lowest dissolved silica concentrations will gibbsite precipitate before
one of the aluminosilicates.
For the most part, silicates do not dissolve in the conventional sense, rather they react with water
to release some ions to solution and form new minerals in place of the original ones. This phenomenon is known as incongruent solution. In considering such reactions, we can usually assume
all Al remains in the solid phase. If we consider only Al and Si, a simple reaction might be the
breakdown of kaolinite to form gibbsite plus dissolved silica:
1
5
Al Si O (OH)4(s) + H2 O ® Al(OH)3(s) + H4 SiO4
6.84
2
2 2 2 5
Since the solid phases can be viewed as pure, the equilibrium constant for this reaction is simply:
K = aH 4 SiO 4 = 10 -4
6.85
-4.4
which tells us that at H2SiO4 activities greater than 10 , kaolinite is more stable than gibbsite. Similar reactions can, of course, be written between other phases, such as kaolinite and pyrophyllite.
Introducing other ions into the system allows the possibility of other phases and other reactions,
e.g.:
KAlSi3 O8 + H+ + 7H2 O ® Al(OH)3 + K+ + 3H4 SiO4
and: KAl3 Si3 O10(OH)2 +
K=
aK + aH 4 SiO 4
aH +
a +
3
H++ 32 H 2 O ® 2 Al2 Si2 O5 (OH)4 + K+ K mus - kao = K
aH +
244
6.86
6.87
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Silica
Amorphous
Log (aK+/aH+)
Quartz
From a series of such reactions and their corresponding equilibrium constant expressions,
we can construct stability diagrams such as the
6
Muscoone in Figure 6.22.
The procedure for
vite
K-fspar
constructing this and similar diagrams is
5
essentially similar to that used in construction
of pe–pH diagrams and is illustrated in Example
4
6.8. In the case of Figure 6.22, we seek an equilibrium constant expression containing the
3
aK + /aH+ ratio and the activity of H 4SiO4. From
Gibbsite
Kaolinite
this expression we determine the slope and
2
intercept, which allows us to plot the
predominance boundary. For example, the
1
Pyroboundary between the kaolinite and muscovite
fields is given by equation 6.87. The
phyllite
equilibrium constant for the reaction is 10 4, so
0 -5
-4
-3
-2
the boundary is a line with slope 0 at log
Log
a
H
SiO
aK + /aH+ = 4. The boundary between gibbsite
4 4
and kaolinite is reaction 6.84, and equation Figure 6.22. Stability diagram for the system
6.85, written in log form, defines the line divid- K2O-Al2O3-SiO2-H2O at 25° C. After Drever (1988).
ing the two regions. This boundary thus plots
as a vertical line at log aH 4 SiO4 = -4.4. The kaolinite – K-feldspar boundary is the reaction:
1
9
+
+
2 Al2 Si2 O5 (OH)4 +K +2H4 SiO4 ®KAlSi3 O8 + H + 2 H2 O
K kao - kfs =
log
aH +
aK + aH 4 SiO 4
aK +
= pK - 2 log aH 4 SiO 4
aH +
The boundary thus plots as a line with a slope of -2 and an intercept equal to the negative of the log
of the equilibrium constant. Boundaries for the remaining fields can be derived similarly.
The fields in Figure 6.22 show the phase that is the most stable of those we considered in constructing the diagram. (Strictly speaking, it does not tell us whether the phase can be expected to
precipitate or not, as this depends on the Al activity; however, because of the low solubility of Al,
an aluminum-bearing phase can be expected to be stable in most instances.) By considering sodium
and calcium as well as potassium, we can construct a 3-dimensional stability diagram, such as that
in Figure 6.23, in a similar manner.
Because many low-temperature reactions involving silicates are so sluggish, equilibrium constants
are generally calculated from thermodynamic data rather than measured.
6.5 Clays and Their Properties
Clays are ubiquitous on the surface of the Earth. Unfortunately, the term clay has two meanings in
geology: in a mineralogical sense it refers to a group of sheet silicates, in another sense it refers to
the finest fraction of a sediment or soil. In addition to clay minerals, fine-grained oxides and
hydroxides are present in the ‘clay fraction’. Clays, in both senses, exert important controls on the
composition of aqueous fluids, both because of chemical reactions that form them and because of
their sorptive and ion exchange capacities. Generally, only clays in the mineralogical sense have
true ion-exchange capacity, where ions in the clay can be exchanged for ions in the surrounding
solution, but oxides and hydroxides can adsorb and desorb ions on their surfaces. We will first
consider the mineralogy of the true clays, then consider their interaction with solution.
245
November 6, 2001
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Chapter 6: Aquatic Chemistry
6.5.1 Clay Mineralogy
6.5.1.1 Kaolinite Group (1:1
Clays)
Anorthite
ca
11
spar
d
l
e
f
K-
10
i
K-m
11
10
9
9
8
8
psite
7
6
Phil
li
Log aK+/aH+
Clay minerals (sensu stricto)
are sheet silicates. We can think
of each sheet in a clay mineral
as consisting of layers of silica
tetrahedra
bound
to
a
hydroxide layer in which the
cation (most commonly Al, Mg,
or Fe) is in octahedral coordination much as in pure
hydroxide minerals such as
gibbsite (Al(OH)3) or brucite
(Mg(OH)2). Brucite and gibbsite
are structurally similar to each
other except that in gibbsite
every third octahedral site is left
empty to maintain charge
balance. This is illustrated in
Figure 6.24. Because only 2 out
of 3 octahedral sites are occupied, gibbsite is said to have a
dioctahedral structure,
while
brucite is said to be trioctahedral.
This terminology also applies in
clay minerals in exactly the
same sense.
5
-8
site
Gibb
-7
-6
Log -5
a
H4 S
iO
4
l
Kao
-4
-3
5
6
7
6
5
ite 11
Alb
10
9
8
+
7 a Na+/a H
Log
Smectite
Figure 6.23. Stability diagram for the system K2O-Na2O-CaOAl2O3-SiO2-H2O at 25° C. After Garrels and Christ (1965).
The simplest clays consist of a
tetrahedral silicate layer and an octahedral hydroxide layer, hence the term 1:1 clays. Kaolinite,
Al4Si4O 10(OH)8, is a good example. The structure of kaolinite is shown in Figure 6.25. Its unit cell
consists of a layer of silica tetrahedra bound to an octahedral alumina layer whose structure is very
similar to that of gibbsite except that some oxygens are replaced by hydroxyls. Note that individual
sheets are not bound together; they are held together only by van der Waals interactions, which are
quite weak. The structure of serpentine, Mg6Si4O10(OH)8, is similar to that of kaolinite, with Mg replacing Al, and every octahedral site is occupied. In both minerals, a mismatch in the spacing of
octahedra and tetrahedra results in a curvature of the lattice. Also successive layers of kaolinite are
generally stacked in a random manner.
6.5.1.2 Pyrophyllite Group (2:1 Clays)
This is a large group of clay minerals consisting of a "hydroxide layer" sandwiched between two
layers of silica tetrahedra, hence the term 2:1 clays. More of the hydroxyls in the hydroxide layer
are replaced by oxygen than in 1:1 clays. The simplest two such clays are pyrophyllite,
Al2Si4 O10(OH)2, and talc, Mg3Si4 O 10(OH) 2. In pyrophyllite, every third octahedral site is vacant, so
that it, like kaolinite, is said to be dioctahedral, while serpentine and talc are trioctahedral. The
structure of pyrophyllite is shown in Figure 6.26. Other, more complex clays, including those of
the smectite group, the biotite group, the vermiculite group, saponite, and muscovite, have structures similar to those of pyrophyllite and talc. These are derived in the following ways (Figure
6.27):
246
November 6, 2001
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Chapter 6: Aquatic Chemistry
Substitution of Al+3 for Si+4 in the tetrahedral sites resulting in a charge deficiency that is balanced by the presence of a cation between the layers
(interlayer positions).
When the
number of interlayer cations is small,
they are generally exchangeable;
when the number is large, K is typically the cation and it is not very exchangeable.
2. Substitution of Mg 2+, Fe 2+ , Fe 3+ or a
similar cation for Al 3+ or substitution
(a)
of Fe 2+ , Fe 3+ , or Al 3+ for Mg 2+ in the
Hydroxyl
octahedral layer. Where charge defi2+
ciency results, it is balanced by the
(3 Mg )
presence of exchangeable cations in
3+
or (Al
the interlayer sites, or by vacancies in
+ vacancy)
the octahedral sites.
Hydroxyl
(b)
Smectites are distinguished by their
expansion
to a unit cell thickness of 14
Mg2+ in brucite
Å upon treatment with ethylene glycol.
vacant in gibbsite
This expansion results from entry of the
Hydroxyl
2+
ethylene glycol molecule into the interMg in brucite
layer
position of the clays. Smectites
3+
Al in gibbsite
generally also have water present in the
Figure 6.24. Structure of gibbsite and brucite. (a) plan
interlayer space, the amount of water
(vertical) view; (b) expanded x-sectional view.
1.
being determined by the cation present. Generally there is little water
present when the interlayer cation
is divalent Mg or Ca, but can be
very considerable when the cation
is Na. The amount of water also
depends on the humidity; as a result
smectites, and sodium bearing ones
in particular will swell on contact
(a)
with water, affecting permeability.
The most common smectite is montmorillonite,
X1/3(Mg1/3Al5/3)Si4O10
.
(OH)2 nH 2O. The interlayer cation
of smectites is exchangeable: in a
Octahedral
NaCl solution it will be Na, in a
7Å
Layer
CaCl2 solution it will be Ca, etc.
Tetrahedral
Vermiculites have a higher net
charge on the 2:1 layer (that is,
Layer
there is greater cation deficiency).
As a result, the electrostatic forces
holding the layers together are
greater so that the interlayer space
O
OH
Al
Si
(b)
is less expandable and the interlayer
cations less exchangeable.
Figure 6.25. Structure of kaolinite. (a) Plan view of the tetrahedral layer. (b) Cross-sectional view.
247
November 6, 2001
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G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
Interlayer positions
Tetrahedral
Layer
9.5 Å
Octahedral
Layer
Tetrahedral
Layer
O
OH
Si
Al (pyrophillite)
or Mg (talc)
Vacancy (pyrophyllite)
or Mg (talc)
Figure 6.26. Structure of pyrophyllite.
smectite or hydroxy-interlayer
smectite.
6.5.1.3 Chlorite Group (2:2
clays)
This group is characterized
by having a unit cell consisting of two tetrahedral layers
and two (hydroxide) octahedral layers. The ideal formula
of chlorite is (Mg,Fe,Al)6
(SiAl)4O10(OH)8, where the
elements in parentheses can
be in any proportions. The
structure of chlorite is shown
in Figure 6.28. Chlorites with
unit cells consisting of a single tetrahedral and single octahedral layer also occur and
are called septechlorite (because they have a 7Å spacing;
true chlorites have 14Å spacing), but are less stable and
uncommon.
Al2Si4O10(OH)2
Pyrophyllite
1 + 2+
3 (X Mg )
for 13 Al3+
1 + 3+
3 (X Al )
for 13 Si4+
Micas (biotite and muscovite)
are related to pyrophyllite and
talc by substitution of an Al3+ for a
Si4+ in a tetrahedral site. The result is that the silicate layers are
relatively strongly bound to the
interlayer K, which is not normally exchangeable.
Illite is a
name applied to clay-sized micas,
though it is sometimes restricted
to the dioctahedral mica (muscovite). Generally, illite has less K
and Al and more Si than igneous
or metamorphic muscovite.
Because of the structural similarity of various 2:1 clays, they can
form crystals that consist of layers
of more than one type, for example illite-smectite.
In addition,
layers of gibbsite or brucite may
occur in smectite. Different layers
may be distributed randomly or
may be ordered. These are called,
for example, mixed-layer chlorite-
X1/3(Mg1/3Al1 2/3)Si4O10(OH)2.nH2O
Montmorillonite
X Al (Si Al )O (OH) .nH O
1/3 2
3 2/3 1/3
10
2
2
Beidellite
KAl2(Si3Al)O10(OH)2
Muscovite
(K+Al3+)
for Si4+
1 + 3+
3 (X Al )
.
Mg3Si4O10(OH)2
1 4+ X1/3Mg3(Si3 2/3Al1/3)O10(OH)2 nH2O
for 3 Si
Saponite
Talc
2 + 3+
3 (X Al ) X Mg (Si Al )O (OH) .nH O
2/3 3 3 1/3 2/3 10
2 2
(K+Al3+) for 23 Si4+
Vermiculite
for Si4+
KMg3(Si3Al)O10(OH)2
Phlogopite
Fe2+ for Mg2+
Biotite
KFe3(Si3Al)O10(OH)2
Annite
Figure 6.27. Relationship of 2:1 clays to pyrophyllite and talc. X
1
1
represents an exchangeable cation (K, Na, 2 Mg, 2 Ca, etc.). From
Drever (1988).
248
November 6, 2001
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Chapter 6: Aquatic Chemistry
6.5.2 Ion-Exchange
Properties of Clays
One of the most important
properties of clays is their capacity for ion exchange. In soil
science, the term ion exchange is
refers specifically to replacement of an ion adsorbed to the
surface by one in solution.
However, we shall use the term
in a more general sense here, 14 Å
and include as well exchange
reactions between ions in solution and ions bound within the
solid. The ability of a substance
to exchange ions is called the
ion exchange capacity and is generally measured in equivalents
or milliequivalents (meq). Ion
exchange capacities of clays are
listed in Table 6.3.
O
OH Si Mg, Fe, Al, or vacancy
The exchange reaction of two
Figure 6.28. Structure of Chlorite.
monovalent ions between clay
and solution may be written:
Tetrahedral
Octahedral
Tetrahedral
Octahedral
Xclay + Y+ ® Yclay + X+
The corresponding equilibrium constant expression is written as:
K=
aY -clay aX +
aX -clay aY +
aY -clay
a +
=K Y
aX -clay
aX +
or:
6.88
If we express this with molar concentrations in solution and mole fractions (X) in the solid rather
than activities:
XY -clay
[Y+ ]
= KYX +
X X -clay
[X ]
K' is called the selectivity constant. It expresses the selectivity of the clay for the Y ion over the X
ion. Because we have expressed it in mole fraction rather than activity, we can expect its value to
depend on the composition of both the clay and the solution. We may also define a distribution coefficient, Kd as:
Kd =
[ Xclay ]
[X + ]
6.89
The Kd is generally applied to trace elements. Where Henry's Law holds, Kd should be independent of the concentration of the ion in solution or in the clay, but it will nevertheless depend on the
overall composition of the solution and the clay (in other words, the Henry's law constant will be
different for different clays and solutions).
A more general expression for the equilibrium constant is:
aYn -clay
anX -clay
=K
aYn
anX
6.90
where n is the stoichiometric coefficient.
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November 6, 2001
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Chapter 6: Aquatic Chemistry
The power term is important. Consider the case of exchange between Na+ and Ca2+. The reaction
is:
2+
+
+
2 Na clay
+ Ca 2aq+ ® Ca clay
+ 2 Na aq
The K' expression is:
XCa 2+
2
X Na
+
=K
[Ca 2+ ]
[Na + ]2
6.91
6.92
If we assume that (1) the mole fractions of Na + and Ca2+ in the clay must sum to one (i.e., they are
the only ions in the exchanging site), (2) molar concentrations in solution of 1 for Na and Ca in solution, and (3) K' = 1, solving equation 6.92 yields XNa = 0.62 and XCa = 0.38. If we kept the ratio of
Ca and Na in solution constant, but dilute their concentrations 1000 fold, we obtain X Na = 0.03 and
XCa = 0.97. Thus by diluting the solution, the divalent cation has almost entirely replaced the
monovalent ion. The composition of the clay will thus depend on the ionic strength of the solution. The dominant exchangeable cation is Ca2+ in fresh water, but Na+ in seawater.
6.6 Mineral Surfaces and Their Interaction With Solutions
Reactions between solutions and solids necessarily involves the interface between these phases.
The details of interface processes thus govern equilibria between solids and solutions. Because
clays and other sedimentary particles are typically very small, their surface area to volume ratio is
high. This adds to the importance of surface chemistry. For example, the concentrations of many
trace elements, particularly the transition metals, dissolved in streams, rivers, and the oceans are
controlled not by precipitation and dissolution, but rather by adsorption on and desorption from
mineral and organic surfaces. These surface reactions maintain the concentrations of these elements
in seawater well below saturation levels. Surface processes also play an important role in soil fertility. Soils have concentrations in soils above levels one would predict from equilibrium dissolution
and precipitation because of adsorption onto particles surfaces. Surface adsorption will also
strongly affect the dispersion of pollutants in soils, ground and surface waters. We discussed some
aspects of surface chemistry in Chapter 5 within the context of kinetic fundamentals. We return to it
in this chapter in a broader context.
6.6.1 Adsorption
We can define adsorption as attachment of an ion in solution to a preexisting solid surface, for example a clay particle. Adsorption involves one or more of the following:
∑ Surface complex formation: The formation of coordinative bonds between metals and ligands at
the surface. Considered in isolation, this process is very similar to the formation of complexes
between dissolved components.
∑ Electrostatic interactions: As we shall see, solid surfaces are typically electrically charged. This
electrostatic force, which is effective over greater distances than purely chemical forces, affects
surface complex formation and loosely binds other ions to the surface.
∑ Hydrophobic adsorption: Many organic substances, most notably lipids, are highly insoluble in
water due to their non-polar nature. These substances become adsorbed to surfaces, not because they are attracted to the surface, but
rather because they are repelled by water.
Table 6.3. Ion Exchange Capacity of
The interaction of the three effects make quantiClays
tative prediction of adsorption behavior more diffiClay
Exchange Capacity (meq/100g)
cult than prediction of complexation in solution.
Smectite
80-150
The functional groups of mineral and organic surVermiculite
120-200
faces have properties similar to those of their disIllite
10-40
solved counterparts. In this sense, surface comKaolinite
1-10
plexation reactions are similar to complexation reChlorite
<10
actions in solution. However, reactions between
250
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
these surface groups and dissolved species is complicated
by the proximity of surface groups to each other, making
them subject to long range electrostatic forces from neighboring groups. For example, the surface charge will
change systematically as the adsorbed surface concentration
of a positive species such as H + increases. This change in
surface charge will decrease the attraction between H + ions
and the surface. As a result, the equilibrium constant for the
surface protonation reaction will change as the surface concentration of H+ increases.
We found in Chapter 5 that adsorption is usually described in terms of adsorption isotherms. We introduced
two such isotherms, the Langmuir isotherm:
K ad [A]
QA =
1 + K ad [A]
(5.127)
d– d+ d– d+ d– d+ d– d+ d–
a
d+
b
d+
d–
d+
d–
d+
d–
d+
d–
d– d+ d– d+ d– d+ d– d+ d–
(where QA is the fraction of surface sites occupied by species
A, [A] is the dissolved concentration of A, and Kad is the adsorption equilibrium constant), and the Freundlich isotherm:
Q A = K ad [A]n
(5.131)
where n is an empirical constant. We derived the Langmuir
isotherm from kinetic fundamentals, though we could have
also derived it from thermodynamics. Inherent in its derivation are the assumptions that (1) the free energy of adsorption is independent of the number of sites available and
therefore that (2) the law of mass action applies and that (3)
only a monolayer of adsorbate can form. The Langmuir
isotherm thus shows a decrease in the fraction of A adsorbed when the concentration of A in solution is high, reflecting saturation of the surface. In contrast, the Freundlich
isotherm, which is merely empirical, shows no saturation.
We also found that at low relative saturation of the surface,
the Freundlich isotherm with n=1 approximates the Langmuir isotherm.
Adsorption phenomena are generally treated with the surface complexation model, which is a generalization of the
Langmuir isotherm (Stumm and Morgan, 1996; Morel and
Hering, 1993). The model incorporates both chemical bonding of solute species to surface atoms and electrostatic interactions between the surface and solute ions. The model assumes
that these two effects can be treated separately. Thus the
free energy of adsorption is the sum of a complexation, or
intrinsic, term and an electrostatic, or coulombic term:
c
Figure 6.29. (a) Metal ions (small red
spheres) and oxygens (large gray
spheres) on a mineral surface are incompletely coordinated, leading to a
partial charge on the surface (indicated by d + and d–). (b) When the
mineral surface is emmersed in water, water molecules coordinate metal
ions on the surface.
(c)
Water
molecules will dissociate leaving hydroxyl groups coordinating metal
ions. Protons (small dark spheres)
will associate with surface oxygens,
forming additional hydroxyl groups.
DGad = DGintr + DGcoul
6.93
From this it follows that the adsorption equilibrium constant can be written as:
Kad = Kintr Kcoul
6.94
Letting ∫S denote the surface site and A denote a solute species, we may write the adsorption reaction as:
∫S + A ® ∫SA
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Chapter 6: Aquatic Chemistry
Let's begin by considering comparatively simple surfaces: those of metal oxides. Although silicates are likely to be more abundant that simple oxides, the properties of silicate surfaces approximate those of mixtures of their constituent oxides surfaces. Hence what we learn from consideration
of oxides can be applied to silicates as well. We will initially focus just on the intrinsic terms in
equations 6.93 and 6.94. We will return to the coulombic term at the end of this section.
Oxygen and metal atoms at an oxide surface are incompletely coordinated; i.e., they are not surrounded by oppositely charged ions as they would be in the interior of a crystal (Figure 6.29a).
Consequently, mineral surfaces immersed in water attract and bind water molecules (Figure 6.29b).
These water molecules can then dissociate, leaving a hydroxyl group bound to the surface metal
ion. We may write this reaction as:
∫M+ + H2 O ® ∫MOH + H+
where ∫M denotes a surface metal ion.
In a similar fashion, incompletely coordinated oxygens at the surface can also bind water molecules, which can then dissociate, again creating a surface hydroxyl group:
∫O– + H2 O ® ∫OH + OH–
Thus the surface on an oxide immersed in water very quickly becomes covered with hydroxyl
groups (Figure 6.29c), which we can write as ∫SOH and which are considered to constitute part of
the surface rather than the solution. These hydroxyl groups can then act as either proton acceptors
or proton donors through further association or dissociation reactions, e.g.:
∫SOH + H+ ® ∫SOH 2+
or
(a)
X – O H + Mz+
X – O – M(z-1) + H+
(b)
X – O H + L-
X – L + OH-
(c)
X – O – M(z-1)+ L-
X – O – M – L(z-2)
(d)
X – L + Mz+
X – L – M(z+1)
(e)
(f)
X–O–H
X–O–H
X–O–H
X–O–H
+ Mz+
+ HPO4
X–O
X–O
M(z-2)++ 2H+
X–O
OH
+ 2OHX–O P O
Figure 6.30. Complex formation of solid surfaces may occur
when (a) a metal replaces a surface proton, or (b) a ligand
replaces a surface OH group. The adsorbed metal (c) may
bind an additional ligand, and the ligand (d) may bind an
additional metal. Multidentate adsorption involves more
than one surface site (e, f).
252
∫SOH ® ∫SO- + H+
We should not be surprised to
find that these kinds of reactions
are strongly pH dependent.
Adsorption of metals to the surface may occur through replacement of a surface proton, as is illustrated in Figure 6.30a, while
ligands may be absorbed by replacement of a surface OH group
(Figure 6.30b).
The adsorbed
metal may bind an additional
ligand (Fig. 6.30c), and the adsorbed ligand may bind an additional metal (Fig. 6.30d).
An additional possibility is
multidentate adsorption, where a
metal or ligand is bound to more
than one surface site (Figures
6.30e and 6.30f). This raises an interesting dilemma for the Langmuir isotherm. Where x sites are
involved, we could write the reaction as:
x∫S + A ® ∫Sx A
and the corresponding equilibrium constant expression as:
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
K ad =
[∫ Sx A]
[∫ S]x [ A ]
6.95
100
80
60
40
20
0
3–
Percent adsorption
percent bound
AsO4
CrO42–
where x is the number of sites involved and A is
2–
the species being adsorbed. This assumes, howSO4
SeO42–
ever, that the probability of finding x sites toVO43–
gether is proportional to the xth power of concentration, which is not the case. A better approach
is to assume that the reaction occurs with a
2 3 4 5 6 7 8 9 10 11 12 13
multidentate surface species, ∫Sx and that its conpH
centration is [∫S]/x. The equilibrium constant is
then:
Figure 6.31. Binding of ligands (anions) on the
surface of hydrous ferric oxide (SFe = 10-3 M)
[∫ Sx A]
K ad =
from dilution solution (5 ¥ 10–7 M; I=0.1) as a
[ A ][∫ S] / x
function of pH. From Stumm and Morgan
Alternatively, the /x can be contained within the (1996).
equilibrium constant.
Since surface bound protons and OH
100
groups are almost inevitably involved
H
E
SiO2 25∞C J
in adsorption, we would expect that
adsorption of metals and ligands will be
J
E
strongly pH dependent. This is indeed
80
J
C
H
the case, as may be seen in Figure 6.31
C
J
C
and 6.32: adsorption of cations increases
J
with increasing pH while adsorption of
50
H
anions decreases with decreasing pH.
Ca(II)
J
H
Co(II)
H
E
E
Figure 6.32 shows that adsorption of
EE
H
H
H
C
Fe(III) Cr(III) E
metals on SiO2 goes from insignificant
C
40
to nearly complete over a very narrow
H
C
range of pH. This strong dependence
E
H
HH
E
on pH certainly reflects protonation of
EE
J
the surface as we have discussed above,
20
H
H
but it also reflects the extent of
C
H
E
J
HH
J
C C
hydrolysis of the ion in solution.
H
H
J
HC
H
H
As is the case with soluble complexes,
C
surface complexes may be divided into
0
2
4
6
8
10
12
inner sphere and outer sphere compH
plexes (Figure 6.34).
Inner sphere Figure 6.32. Adsorption of metals on SiO2. Symbols show
complexes involve some degree of co- experimentally determined values; curves show theoretical
valent bonding between the adsorbed values (from James and Healy, 1972).
species and atoms on the surface. In an
outer-sphere complex, one or more water molecules separate the adsorbed ion and the surface; in
this case adsorption involves only electrostatic forces. The third possibility is that an ion may be
held within the diffuse layer (see following section) by long range electrostatic forces.
6.6.2 Development of Surface Charge and the Electric Double Layer
Mineral surfaces develop electrical charge for three reasons:
1. Complexation reactions between the surface and dissolved species, such as those we discussed in
the previous section. Most important among these are protonation and deprotonation. Because
these reactions depend on pH, this aspect of surface charge is pH dependent. This pH dependence
is illustrated in Figure 6.35.
253
November 6, 2001
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Chapter 6: Aquatic Chemistry
Example 6.9. Adsorption of Pb2+ on Hydrous Ferric Oxide as a Function of pH
Using the following apparent equilibrium constants:
∫FeOH 2+ ® ∫FeOH + H+
pKa 1 = 7.29
6.96
∫FeOH ® ∫FeO– + H+
pKa 2 = 8.93
6.97
∫FeOPb+ ® ∫FeO– + Pb2 + pKad = 8.15
Pb2 + + H2 O ® PbOH+ + H+
pKO H = 7.7
calculate the fraction of surface adsorbed Pb as a function of pH from pH 5 to pH 8 for
concentrations of surface sites of 10-3 M, 10-4 M, and 10-5 M assuming a total Pb concentration of
10-9 M.
Answer: The quantity we wish to calculate is [∫FeOPb +]/SPb, so we want to find an expression
for [∫FeOPb + ] as a function of pH. We chose our components to be H+ , Pb2+ , and ∫FeOH +2 and
begin by writing the two relevant conservation equations:
SPb = [Pb2 +] + [PbOH+] + [∫FeOPb+]
+
2
6.98
S∫Fe = [∫FeOH ] + [∫FeOH] + [∫FeO ] + [∫FeOPb ]
–
+
6.99
From the equilibrium constant expressions, we have the following:
[FeOH+2 ]
[FeOH] =
K a1
[H + ]
[FeOH]
[FeOH 2+ ]
[FeO - ] =
K
=
K a1K a 2
a2
[H + ]
[H + ]2
[FeOH+2 ]
K K
+
2+
-1
[FeOPb ] = [FeO ][Pb ]K ad =
[Pb 2 + ] a1 a 2
+ 2
K ad
[H ]
6.100
6.101
6.102
Substituting equations 6.100-6.102 into 6.99 we have:
Ï
K
K K
[Pb 2+ ] K K ¸
S ∫ Fe = [FeOH +2 ]Ì1 + a+1 + a1 + a2 2 + + 2 a1 a 2 ý
K ad þ
[H ]
Ó [H ] [H ]
6.103
Since the [Pb2+] is small, the last term on the right can be neglected so we have:
Ï
K
K K ¸
S ∫ Fe @ [FeOH +2 ]Ì1 + a+1 + a1 + a2 2 ý
Ó [H ] [H ] þ
In a similar way, we obtain:
Ï [FeOH+2 K a1K a 2 KOH ¸
S ∫ Pb = [Pb 2+ ]Ì1 +
+ + ý
[H + ]2 [H + ]2
[H ] þ
Ó
Solving this pair of equations for [∫FeOH +2 ] and [Pb2+], and substituting these into 6.102, we obtain:
¸
Ï
ÔÊ
Ô
[FeOH ]
SPb
Ô K a1K a 2 ˆ
Ô
[FeOPb + ] = + 2
ý
Ì
+
KOH ÔÁË K ad ˜¯
[H ] + [H ]K a1 + K a1K a 2 Ô S ∫ Fe K a1K a 2 / K ad
+
ÔÓ [H + ]2 + [H+ ]K a1 + K a1K a 2 [H+ ] Ôþ
+
2
Dividing by SPb and simplifying, we have:
[FeOPb + ]
S ∫ FeKa2
=
+
+ 2
SPb
K ad ([H ] / K a1 + [H ] + K a 2 ) + ([H+ ] / K a1 + 1 + K a 2 / [H+ ])KOH K ad + S ∫ FeKa2
The result is shown in Figure 6.33. For the highest concentration of surface sites, Pb goes from
virtually completely in solution to virtually completely adsorbed within 2 pH units.
254
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
M
10 -5
6
1¥
5.5
0 -4 M
1¥1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
05
1 ¥1
0 -3 M
fraction Pb adsorbed
Chapter 6: Aquatic Chemistry
6.5
pH
7
7.5
8
2. Lattice imperfections at the solid surface
as well as substitutions within the crystal
lattice (e.g., Al 3+ for Si4+). Because the
ions in interlayer sites of clays are readily
exchangeable, this mechanism is particularly important in the development of surface charge in clays.
3. Hydrophobic adsorption, primarily of
organic compounds, and “surfactants” in
particular. We will discuss this effect in
Chapter 14.
Thus there are several contributions to
surface charge density. We define snet as
the net density of electric charge on the solid
surface, and can express it as:
snet = s0 + sH + sSC
Figure 6.33. Calculated adsorption of Pb 2+ on hydrous
ferric oxide for three different concentrations of surface
sites: 10-3 M, 10-4M, and 10-5 M.
Outer Sphere Complex
6.104
where s0 is the intrinsic surface charge
due to lattice imperfections and substitutions, sH is the net proton charge,
Diffuse Layer Ion
i.e., the charge due to binding H +
and OH –, s SC is the charge due to
other surface complexes. s is usually measured in coulombs per
square meter (C/m2). sH is given
Inner Sphere Complex by:
sH = F(GH – GOH)
6.105
where F is the Faraday constant
and GH and GOH are the adsorption
densities (mol/m2) of H + and OH –
respectively. In a similar way, the
charge due to other surface complexes is given by
sSC = F(zM GM + zA GA ) 6.106
Figure 6.34. Inner sphere surface complexes involve some
degree of covalent bonding between the surface and the ion;
outer sphere complexes form when one or more water molecules intervenes between the surface and the ion. Ions may
also be held in the diffuse layer by electrostatic forces.
where the subscripts M and A refer
to metals and anions respectively,
G is again adsorption density and z
is the charge of the ion. The surface complex term may also be
broken into an inner sphere and outer sphere component:
sSC = sIS + sOS
6.107
Thus net charge on the mineral surface is:
snet = s0 + F(GH – GOH + ZM GM + ZA GA )
6.108
Figure 6.35 shows that at some value of pH the surface charge will be zero. The pH at which this
occurs is known as the isoelectric point, or zero point of charge (ZPC). The ZPC is the pH at which the
charge on the surface of the solid caused by binding of all ions is 0, which occurs when the charge due
to adsorption of cations is balanced by charge due to adsorption of anions. A related concept is the
point of zero net proton charge (pznpc), which is the point of zero charge when the charge due to the
binding of H+ and OH– is 0; i.e., s H = 0. Table 6.4 lists values of the point of zero net proton condi255
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
tion for some important solids.
Surface charge depends on the nature of the surface, the nature of
the solution, and the ionic
strength of the latter. An important feature of the point of zero
charge, however, is that it is independent of ionic strength, as is illustrated in Figure 6.36.
Surface Charge C/m2
rs
Se r
–0.2
0.2
0.1
Hy Alg
dr
in
0
–0.1
m
albu
um
Electrophoretic Mobility
mm/s • cm/v
3
ite
0
–0.1
Eff oil dr
luen opl
t ets
2
4
–0.2
Aerobact
er
6
pH
aerogenes
8
10
12
Figure 6.35. (a) Surface charge of some common sedimentary
materials as a function of pH. (b) Electrophoretic mobility, which
is related to surface charge, of representative organic substances
as a function of pH. The pH dependence of surface charge reflects the predominance of attached protons at low pH and the
predominance of attached hydroxyls at higher pH. From Stumm
(1992).
where CA and CB are the concentrations of conjugate of the acid
or base added (e.g., Na + is the conjugate of the base NaOH)
+
and [∫SOH 2 ] and [∫SO–] are the concentrations (in moles per liter) of the surface species. The surface charge, Q (in units of
moles of charge per liter), is simply:
Q= [∫SOH 2+ ] – [∫SO– ]
So that the surface charge is determined from:
Q = C A - C B - [H + ] +
10 -14
[H + ]
6.109
The surface charge density, s, is calculated from Q as:
s=
Cal g Al
c 2O
O
Mg
a
Ka M
2
CA – CB + [OH– ] - [H+] =
[∫SOH 2+ ] – [∫SO– ]
Mon
tmo
ae bon
r
oca age
Se w
The surface charge due to binding of protons and hydroxyls is
readily determined by titrating a
solution containing a suspension
of the material of interest with
strong acid or base. The idea is
that any deficit in H + or OH – in
the solution is due to binding
with the surface. For example,
consider a simple hydroxide sur+
face with surface species ∫SOH 2
–
0
and ∫SO (as well as ∫SOH ).
Charge balance requires that:
0.1
nO ite
a
olin , Feldsp
te
oni
rill SiO 2
6.6.2.1 Determination of Surface
Charge
0.2
QF
A[∫ S]
6.110
Table 6.4. Point of Zero Net
Proton Charge of Common
Sedimentary Particles
Material
SiO2 (quartz)
SiO2 (gel)
a-Al2O3
Al(OH)2 (gibbsite)
TiO2 (anatase)
Fe3O4 (magnetite)
a-Fe2O3 (hematite)
FeO(OH) (goethite)
Fe2O3. nH 2O
d-MnO
b-MnO
Kaolinite
Montmorillonite
where A is the specific surface area (m2/mol) and [∫S] is the
concentration of solid (in mols/l)*.
We can write equilibrium constant expressions for the surface
protonation and deprotonation reactions. For example, for sur- From Stumm (1992).
face protonation:
pH
2.0
1.0-2.5
9.1
8.2
7.2
6.5
8.5
7.8
8.5
2.8
7.2
4.6
2.5
∫SOH + H+ ® ∫SOH 2+
* If the concentration of solid is expressed in kg/l, as it commonly is, then the specific surface area should be in
units of m 2 /kg.
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November 6, 2001
W. M. Wh i t e
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Chapter 6: Aquatic Chemistry
the equilibrium constant is:
30
20
I=0.1 M
0.01 M
0.001 M
0.1
K=
10
0
0
s (mC cm-2)
Q (mM = [FeOH2+– FeO–]
0.2
-10
-0.1
-20
4 5
6 7
6.111
We may write a conservation equation for the
surface as:
S∫S = [∫SOH 2+ ] + [∫SO– ] + [∫SOH0 ] 6.112
At pH below the pzpc, we can consider the entire
surface charge as due to [∫SOH +2 ], so that Q ª
[∫SOH +2 ], and [∫SO–] ª 0. Combining equations
6.109, 6.111 and 6.112, we have:
8 9 10 11
pH
Figure 6.36. Surface charge on FeOOH as a
function of pH for different ionic strengths of
a 1:1 electrolyte (10-3 M FeOOH). From
Dzombak and Morel (1990).
[∫ SOH 2+ ]
[∫ SOH][H+ ]
K=
Q
( S ∫ S - Q )[H+ ]
6.113
In equation 6.109, we see that if the amount of acid
(or base) added is known, the surface charge can
be determined by measuring pH (from which
[OH–] may also be calculated). This is illustrated in
Figure 6.37.
Thus the value of the protonation reaction equilibrium constant may be calculated from the surface charge and pH. The equilibrium constant for the deprotonation reaction may be obtained in a
similar way. These equilibrium constants are also known as surface acidity constants, and sometimes
denoted (as in Example 6.8) as K a1 and K a2 for the protonation and deprotonation reaction respectively.
6.6.2.2 Surface Potential and the Double Layer
The charge on a surface exerts a force on ions in the adjacent solution and gives rise to an electric
potential, Y (measured in volts), which will in turn depend on the nature and distribution of ions in
solution, as well as intervening water molecules. The surface charge results in an excess concentration of oppositely charged ions (and a deficit of liked charged ions) in the immediately adjacent solution.
The surface charge, s, and potential, Y 0, can be related by Gouy-Chapman Theory‡ , which is
conceptually and formally similar to Debye–Hückel Theory (Chapter 3). The relationship is given
by:
Ê zY F ˆ
s = (8 RTee0 I )1/ 2 sinh Á 0 ˜
Ë 2 RT ¯
6.114
where z is the valance of a symmetrical background electrolyte (e.g., 1 for NaCl), Y 0 is the potential
at the surface, F is the Faraday constant, T is temperature, R is the gas constant, I is ionic strength, e
is the permittivity of water and e0 is the permittivity of a vacuum (see Chapter 3). At 25°C, equ.
6.114 may be written as:
s = aI 1/ 2 sinh (bzY0 )
6.115
where a and b are constants with value of 0.1174 and 19.5 respectively.
Where the potential is small, the potential as a function of distance from the surface is:
‡ Gouy-Chapman Theory assumes a infinite flat charge plane in one dimension. The electrostatic interaction
between the surface and a cloud of charged particles is described by the Poisson-Boltzmann equation, as in
Debye-Hückel Theory. Unlike Debye-Hückel, the Poisson-Boltzmann equation has an exact case. The theory was
developed by Gouy and Chapman around 1910, a decade before Debye and Hückel developed their theory. See
Morel and Herring (1993) for the details of the derivation.
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Chapter 6: Aquatic Chemistry
Y(x) = Y0 e- kx
a
9
where k has units of inverse length and is called the Debye parameter or Debye length and is given by:
pH 7
k=
5
9
1.0
0.5
0
0.5
1.0
mmol/liter
CNaOH
CHClO4
b
6.117
From equation 6.116, we see that the inverse of k is the
distance at which the electrostatic potential will decrease
by 1/e.
The variation in potential, the Debye length,
and the excess concentration of counter-ions with distance from the surface is illustrated in Figure 6.38.
An addition simplification occurs where the potential is
small, namely that equation 6.114 reduces to:
s @ ee0 kI2/3Y0
pH 7
+10
–5
0
+5
Q , 10-5 mol/g
–10
Figure 6.37. (a). Titration of a suspension of a-FeOOH (goethite) (6 g/liter)
by HClO4 and NaOH in the presence
of 0.1 M NaClO4. (b). Charge calculated by charge balance (equation
6.109) from the titration. From Stumm
(1992).
ered to be part of the solid rather than
the Stern Layer. The thickness of the
Gouy (outer) Layer is considered to be
the Debye length, 1/k. As is apparent
in equation 6.116, this thickness will
vary inversely with the square of ionic
strength. Thus the Gouy Layer will collapse in high ionic strength solutions
and expand in low ionic strength ones.
When clays are strongly compacted,
the Gouy layers of individual particles
overlap and ions are virtually excluded
from the pore space. This results in retardation of diffusion of ions, but not of
water. As a result, clays can act as semipermeable membranes.
Because some
ions will diffuse easier than others, a
chemical fractionation of the diffusing
fluid can result.
At low ionic strength, the charged
layer surrounding a particle can be
strong enough to repel similar particles
with their associated Gouy layers. This
As is illustrated in Figure 6.38, an excess concentration
of oppositely charged ions develops adjacent to the surface. Thus an electric double layer develops adjacent to the
mineral surface. The inner layer, or Stern Layer, consists
of charges fixed to the surface, the outer diffuse layer, or
Gouy Layer, consists of dissolved ions that retain some
freedom of thermal movement. This is illustrated in Figure 6.39. The Stern Layer is sometimes further subdivided into an inner layer of specifically adsorbed ions
(inner sphere complexes) and an outer layer of ions that
retain their solvation shell (outer sphere complexes),
called the inner and outer Helmholz planes respectively.
Hydrogens adsorbed to the surface are generally consid-
18
Negatively Charged Surface
Concentration (mM)
5
3
2F 2I
ee0 RT
16 Na+
14
12
1
k
10
2 Cl- 4
8
6
Y
6
10
4
14
2
18
x (nm)
8
12
16
20
Y (mV)
3
6.116
Figure 6.38. Variation in electrical potential and ions with
distance from a negatively charged surface based on the
Gouy-Chapman model. Electrical potential varies exponentially with distance, as do ion concentrations. 1/k is
the distance where the potential has decreased by 1/e.
From Morel and Hering (1993).
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Chapter 6: Aquatic Chemistry
Solid Surface
Layer of
fixed
ions
–
–
–
–
–
–
–
–
–
Electric Potential
Diffuse Layer
–
+
–
–
–
–
–
–
+
–
+
–
–
–
+
–
+
–
–
–
+
–
–
–
Stern
will prevent particles from approaching closely and
hence prevent coagulation. Instead, the particles
form a relatively stable colloidal suspension. As the
ionic strength of the solution increases, the Gouy
layer is compressed and the repulsion between particles decreases. This allows particles to approach
closely enough that they are bound together by attractive van der Waals forces between them. When
this happens, they form larger aggregates and settle
out of the solution. For this reason, clay particles
suspended in river water will flocculate and settle
out when river water mixes with seawater in an estuary.
6.6.2.3 Effect of the Surface Potential on
Adsorption
Gouy
The electrostatic forces also affect complexation
reactions at the surface, as we noted at the beginning of this section. An ion must overcome the
electrostatic forces associated with the electric double layer before it can participate in surface reactions. We can account for this effect by including it
in the Gibbs Free Energy of reaction term, as in
equation 6.93:
1/k
DGads =DGintr+DGcoul
Distance from Surface
Figure 6.39. The double layer surrounding clay particles.
-3.0
taking place in solution),
DG coul is the free energy
to the electrostatic forces
is given by:
H
FeO
6.118
where DZ is the change in molar charge of the surface species due to the adsorption reaction. For example, in the reaction:
∫SOH+Pb2+ ® ∫SOPb+ +H+
The value of DZ is +1 and
DG coul = FY.
Thus if we can calculate
DG coul , this term can be added
to the intrinsic DG for the adsorption reaction (DGintr) to obtain the effective value of D G
(DGads ). From DG ads it is a
simple and straightforward
-5.0
H
FeO
FeO–
-6.0
FeO –
DG = F DZY0
FeOH+2
-4.0
log C
tion
and
due
and
(6.93)
where DG ads is the total free energy of the adsorption reaction, DGint is the intrinsic free energy of the
reaction (i.e., the value the reaction would have in
the absence of electrostatic forces; in general this
will be similar to the free energy of the same reac-
-7.0
-8.0
3
4
5
6
7
pH
8
9
10
11
Figure 6.40. Surface speciation of hydrous ferric oxide for I = 0.1 M
calculated in Example 6.9. Solid lines show speciation when surface
potential term is included, dashed lines show the calculated speciation with no surface potential.
259
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Chapter 6: Aquatic Chemistry
matter to calculate Kads . From equation 3.86 we have:
K = e –DGads/RT
Substituting equation 6.94, we have:
Example 6.10. Effect of Surface Potential on Surface Speciation of Ferric Oxide
Using the surface acidity constants given in Example 6.9, calculate the surface speciation of
hydrous ferric oxide as a function of pH in a solution with a background electrolyte concentration of I
= 0.1 M. Assume the concentration of solid is 10-3 mol/l, the specific surface area is 5.4 ¥ 104 m2/mol
and that there are 0.2 mol of active sites per mole of solid.
Answer: The concentration of surface sites, S∫Fe, is 0.2 mol sites/mol solid ¥ 10-3 mol solid/l = 2 ¥
-4
10 mol sites/l. Our conservation equation is:
S∫Fe = [∫FeOH 2+ ] + [∫FeOH] + [∫FeO– ] = 2 ¥ 10-4
- FDZY / RT
0
P=e
We define P as:
so that our equilibrium constant expressions (6.96 and 6.97) become:
[FeOH+2 ]
K a1P -1
+
[H ]
6.120
[FeOH]
[FeOH+2 ]
-1
K
P
=
K a1K a 2 P -2
a2
+
+ 2
[H ]
[H ]
6.121
[FeOH] =
and
[FeO - ] =
6.119
+
Substituting into our conservation equation, and solving for [∫FeOH 2 ] we have:
-1
Ï K P -1 K K P -2 ¸
[FeOH ] = S ∫ Fe Ì1 + a1 + + a1 +a 2 2 ý
[H ]
[H ]
Ó
þ
The concentration of surface charge, Q, is simply:
Q = [∫FeOH 2+ ] – [∫FeO– ]
F
[FeOH+2 ] - [FeO - ]
and the surface charge density is:
s=
A[ S ]
+
2
(
6.122
)
(P enters the equations as the inverse because we have defined the equilibrium constants in Example
6.9 for the desorption reactions.) Substituting into the surface charge density equation, we have:
Ï K a1K a 2 P -2 ¸
F
+
[
FeOH
]
s=
ý
2 Ì1 A[ S ]
[H+ ]2 þ
Ó
(
)
Substitution equation 6.118 into this, we have:
-1
Ï K a1P -1 K a1K a 2 P -2 ¸ Ï K a1K a 2 P -2 ¸
F
s=
+
S ∫ Fe Ì1 +
ý Ì1 ý
A[ S ]
[H+ ]
[H+ ]2 þ Ó
[H+ ]2 þ
Ó
Finally, substituting equation 6.115 for s, and 6.119 for P, we have:
sinh(bzY0 ) =
F
A[ S ]aI 1/ 2
K a1K a 2 e 2 FDZY0 / RT
1[H+ ]2
S ∫ Fe
Ï K a1e FDZY0 / RT K a1K a 2 e 2FDZY0 / RT ¸
+
Ì1 +
ý
[H+ ]
[H+ ]2
Ó
þ
A pretty intimidating equation, and one with no direct solution. It can, however, be solved by
indirect methods (i.e, iteratively) on a computer. A quick an easy way is to use the Solver feature in
Microsoft Excel™. Figure 6.40 shows the results and compares them to the surface speciation when
surface potential is not considered. The effect of including the surface potential term is reduce the
+
surface concentration of ∫FeO– and ∫FeOH 2 and broaden the pH region where ∫FeO dominates.
260
November 6, 2001
W. M. Wh i t e
G e o c h e mi s t r y
Chapter 6: Aquatic Chemistry
1
and DG coul =
K = Kintr e - FDZY0 / RT
6.123
0.8
l
Since K intr = e
FDGY 0, we have:
6.122
fraction Pb adsorbed
DG intr/RT
RT
poten
t ia
K = e -DG intr / RT e DG coul
ntial
e pote
surfa
c
w it h
w it
hout
surfa
ce
Thus we need only find the value of Y 0,
0.6
which we can calculate from s using
equation 6.114. Example 6.10 illustrates
the procedure.
0.4
The effect of surface potential on a
given adsorbate will be to shift the ad0.2
sorption curves to higher pH for
cations and to lower pH for anions.
Figure 6.41 illustrates the example of
0
adsorption of Pb on hydrous ferric ox4.5
5.5
6.5
7.5
ide. When surface potential is considpH
ered, adsorption of a given fraction of
Pb occurs at roughly 1 pH unit higher Figure 6.41. Comparison of calculated adsorption of Pb
than in the case where surface potential on hydrous ferric oxide with and without including the
is not considered. In addition, the effect of surface potential.
adsorption curves become steeper.
References and Suggestions for Further Reading
Chou, L. and R. Wollast. 1985. Steady-state kinetics and dissolution mechanisms of albite. Amer. J.
Sci. 285: 965-993.
Drever, J. I., 1988. The Geochemistry of Natural Waters, Prentice Hall, Englewood Cliffs, 437 p.
Garrels, R. M. and C. L. Christ. 1965. Solutions, Minerals and Equilibria. New York: Harper and
Row.
James, R. O. and T. W. Healy. 1972. Adsorption of hydrolyzable metal ions at the oxide-water interface. J. Colloid Interface Sci., 40: 42-52.
Morel, F. M. M. and J. G. Hering. 1993. Principles and Applications of Aquatic Chemistry. New York:
John Wiley and Sons.
Pankow, J. F. 1991. Aquatic Chemistry Concepts. Chelsea, MI: Lewis Publishers, 673 p.
Richardson, S. M. and H. Y. McSween. 1988. Geochemistry: Pathways and Processes, New York:
Prentice Hall.
Schlesinger, W. H. 1991. Biogeochemistry. San Diego: Academic Press.
Sposito, G. 1989. The Chemistry of Soils. New York: Oxford University Press.
Stumm, W. 1992. Chemistry of the Solid-Water Interface,. New York: Wiley Interscience 428 p.
Stumm, W. and J. J. Morgan. 1995. Aquatic Chemistry, New York: Wiley and Sons.
White, A. F. and S. L. Brantley (ed.), 1995, Chemical Weathering Rates in Silicate Minerals, Washington: Mineral. Soc. Am.
Problems
1. Using the composition given in Problem 3.4, calculate the alkalinity of seawater at 25°C.
2. For a sodium carbonate solution titrated with HCl to the bicarbonate equivalence point, show
that:
SCO2 = [Cl– ] + [OH+]
3. Calculate the pH of a solution containing SCO2 = 10-2 at 25°C at the bicarbonate and carbonate
equivalence points. Assume ideality and use the equilibrium constants in Table 6.1.
261
November 6, 2001
W. M. Wh i t e
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Chapter 6: Aquatic Chemistry
4. Consider a 0.005M solution of Na2CO3 at 25°C. Assuming ideality:
a. What is the pH of this solution?
b. What is the pH of this solution when titrated to the bicarbonate equivalence point?
c. What is the pH of this solution when titrated to the CO2 equivalence point?
5. Consider a 0.01M solution of NaHCO3 (sodium bicarbonate) at 20°C. Assuming ideality:
a. What is the pH of this solution?
b. Plot the titration curve for this solution (i.e., moles of HCl added vs. pH).
c. What is the pH of the CO2 equivalence point of this solution?
6. Explain why pH changes rapidly near the bicarbonate and CO2 equivalence points during titration.
7. Mars probably once had a more substantial atmosphere and water on its surface. Suppose that it
had a surface atmospheric pressure of 1 bar (0.1 MPa) and that the partial pressure of CO2 was the
same as it is today, 6 ¥ 10-3. Further suppose the surface temperature was 5° C. Use the equilibrium
constant is Table 6.1 for this problem. Under these conditions, at what concentration of Ca2+ ion
would an ancient Martian stream become saturated with CaCO3? What would the pH of that
stream be?
8. Calculate the buffer capacity of a solution initially in equilibrium with calcite for pH between 6
and 9 at 25°C.
9. Calculate the calcium ion concentration for a solution in equilibrium with calcite and fixed SCO2
of 10-2 M at 25°C.
10. Show that the a, fraction of copper complexed as CuOH+ as defined in 6.48, will decrease with
increasing concentration copper in solution. Assume ideal behavior and that H + , OH –, Cu+ and
CuOH+ are the only ions present in solution and that the stability constant of CuOH+ is 10–8.
11. Using the following equilibrium constants and reactions, make a plot of Zn 2+, ZnOH +, Zn(OH)2,
Zn(OH) 3– , Zn(OH) 24 – , and total zinc concentration as a function of pH from pH 1 to pH 14.
Assume ideal behavior and that H + , OH –, and various species of Zn are the only ions in solution.
Hint: Use a log scale for the Zn concentrations.
ZnO +2H+ ® Zn2+ + H2 O
Zn2+ + H2 O ® ZnOH+ +H+
Zn2+ + 2H2 O ® Zn(OH)2 + 2H+
Zn2+ + 3H2 O ® Zn(OH) 3– + 3H+
log Kzn = 11.2
log K1 = –9
log K1 = –16.9
log K1 = –28.1
Zn2+ + 4H2 O ® Zn(OH) 24 – + 4H+
log K1 = –40.2
12. For the adsorption of Zn2+ on hydrous ferric oxide:
∫FeOH0 + Zn2+ ® ∫FeOZn+ + H+
the apparent equilibrium constant is 100.99 . For this problem, use the surface acidity constants (i.e.,
equilibrium constants for adsorption and desorption of H +) given in equations 6.96 and 6.97
(Example 6.9).
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November 6, 2001
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Chapter 6: Aquatic Chemistry
a. Make a plot of Q Zn (fraction of sites occupied by Zn) vs. the aqueous concentration of Zn 2+ at
pH 7 and a total concentration of surface sites of 10-3 M. Assume that Zn 2+ forms no complexes
in solution.
b. Ignoring electrostatic effects and any aqueous complexation of Zn 2+, make a plot of the
fraction of Zn 2+ adsorbed as a function of pH (beginning at pH 6), assuming a total Zn2+
concentration of 10-8M.
c. Do the same calculation as in b, but take into consideration the aqueous complexation
reactions and equilibrium constants in Problem 11.
263
November 6, 2001